Motion in a Straight Line Multiple Choice Question And Answers

Motion in a Straight Line

Question 1. If the velocity of a particle is v = At+Bt2, where A and B are constants, the distance traveled by it between 1 s and 2 s is

  1. \(\frac{3}{2} A+4 B\)
  2. \(\frac{A}{2}+\frac{B}{3}\)
  3. 3A + 7B
  4. \(\frac{3}{2} A+\frac{7}{3} B\)

Answer: 4. \(\frac{3}{2} A+\frac{7}{3} B\)

Given that the velocity at the time t is

⇒ \(v=A t+B t^2 \text { or } \quad \frac{d x}{d t}=A t+B t^2\)

Integrating,

⇒ \(\int d x=\int\left(A t+B t^2\right) d t\)

or, \(x=\frac{A t^2}{2}+\frac{B t^3}{3}+c\)

At t=1s, \(x_1=\frac{A}{2}+\frac{B}{3}+c\)

and at t = 2s, x2 =2A + \(\frac{8}{3}\)B + c.

∴ The distance covered in the given time interval is

⇒ \(x_2-x_1=\left(2 A+\frac{8}{3} B+c\right)-\left(\frac{A}{2}+\frac{B}{3}+c\right)=\frac{3}{2} A+\frac{7}{3} B\)

Question 2. The motion of a particle along a straight line is described by the equation x =8 +12t-t2, where x is in meters and t is in seconds. The retardation of the particle when its velocity becomes zero is

  1. 6ms-2
  2. zero
  3. 12 m s-2
  4. 24 m s-2

Answer: 3. 12 m s-2

The equation representing the position (x) as a function of time (t) is

x = 8 +12t-t3.

Hence, the instantaneous velocity is

⇒ \(v=\frac{d x}{d t}=12-3 t^2\)

and the instantaneous acceleration

⇒ \(a=\frac{d v}{d t}=-6 t\)

When = 0, 12- 3t2 = 0 and t = 2 s.

Hence, the magnitude of the acceleration at t = 2 s is

⇒ \(|a|_{t=2 s}=6(2) \mathrm{ms}^{-2}=12 \mathrm{~m} \mathrm{~s}^{-2} \text {. }\)

∴ retardation =12 ms-2

Question 3. Preeti reached a metro station and found that the escalator was not working. She walked up the stationary escalator in a time tv On other days, if she remains stationary on the moving escalator, the escalator takes her up in a time t2. The time taken by her to walk up the moving escalator is

  1. t1 – t2
  2. \(\frac{t_1 t_2}{t_1+t_2}\)
  3. \(\frac{t_1 t_2}{t_2-t_1}\)
  4. \(\frac{t_1+t_2}{2}\)

Answer: 2. \(\frac{t_1 t_2}{t_1+t_2}\)

Let the velocity of Preeti on the stationary escalator be u0 and the length of the escalator be s. Hence,

⇒ \(u_0=\frac{s}{t_1}\) ….(1)

When the escalator is moving with a velocity ue,

\(u_{\mathrm{e}}=\frac{s}{t_2}\) …(2)

Finally, Preeti’s resultant velocity relative to the ground when both are moving will be u0 + ue. So,

⇒ \(u_0+u_{\mathrm{e}}=\frac{s}{T}\) …(3)

Adding (1) and (2),

⇒ \(u_0+u_{\mathrm{e}}=s\left(\frac{1}{t_1}+\frac{1}{t_2}\right)\)

Equating (3) and (4),

⇒ \(\frac{s}{T}=s\left(\frac{t_1+t_2}{t_1 t_2}\right)\)

Hence, the time T needed for Preeti to walk up the moving escalator is

⇒ \(T=\frac{t_1 t_2}{t_1+t_2}\)

Question 4. The displacement x (in meters) of a particle of mass m (in kilograms) moving in one dimension under the action of a force is related to the time t (in seconds) by t = √x + 3. The displacement of the particle when its velocity is zero will be

  1. 2 m
  2. Zero
  3. 4 m
  4. 6 m

Answer: 2. Zero

Time (t) as a function of position (x) is t = √x +3 or √x =t- 3.

Squaring, x = t²- 6t + 9 = (t- 3)².

∴ instantaneous velocity, \(v=\frac{d x}{d t}=2 t-6\)

For 0 = 0, 2t = 6 or t = 3.

Hence, the position at t = 3 will be x =(t- 3)² = 0.

Question 5. Two cars P and Q start from a point at the same time in a straight line, and their positions are represented by \(x_{\mathrm{P}}(t)=a t+b t^2\) and \(x_{\mathrm{Q}}(t)=f t-t^2\) respectively. At what time do the cars have the velocity?

  1. \(\frac{a+f}{2(1+b)}\)
  2. \(\frac{a+f}{2(b-1)}\)
  3. \(\frac{a-f}{1+b}\)
  4. \(\frac{f-a}{2(1+b)}\)

Answer: 4. \(\frac{f-a}{2(1+b)}\)

For the car P, xp(t) = at + bt², and for the car Q, xQ(t) =ft- f².

∴ The velocity of P is

⇒ \(v_{\mathrm{P}}=\left(\frac{d x}{d t}\right)_{\mathrm{P}}=a+2 b t\)

and the velocity of Q is

⇒ \(v_{\mathrm{Q}}=\left(\frac{d x}{d t}\right)_{\mathrm{Q}}=f-2 t\)

At the time t, let P and Q have the same velocity. So,

Vp = VQ

or, a + 2bt = t – 2t or 2t(b + 1) = f-a.

∴ \(t=\frac{f-a}{2(1+b)}\)

Question 6. A particle covers half its journey with a speed vl and the other half with a speed v2. Its average speed during the complete journey is

  1. \(\frac{v_1 v_2}{v_1+v_2}\)
  2. \(\frac{2 v_1 v_2}{v_1+v_2}\)
  3. \(\frac{v_1+v_2}{2}\)
  4. \(\frac{v_1^2 v_2^2}{v_1^2+v_2^2}\)

Answer: 3. \(\frac{v_1+v_2}{2}\)

Let s be the total distance covered.

For the first half,

⇒ \(\frac{s}{2}=v_1 t_1 \Rightarrow t_1=\frac{s}{2 v_1}\)

Similarly, in the second half,

⇒ \(\frac{s}{2}=v_2 t_2 \Rightarrow t_2=\frac{s}{2 v_2}\)

Hence, the average velocity is

⇒ \(v_{\mathrm{av}}=\frac{\text { total displacement }}{\text { total time taken }}=\frac{s}{t_1+t_2}=\frac{s}{\frac{s}{2}\left(\frac{1}{v_1}+\frac{1}{v_2}\right)}=\frac{2 v_1 v_2}{v_1+v_2} \text {. }\)

Question 7. A particle starts moving from rest with a uniform acceleration of (4/3) m s-2. The distance traveled by the particle during the third second of its motion is

  1. 4 m
  2. 6 m
  3. \(\frac{10}{3} \mathrm{~m}\)
  4. \(\frac{19}{3} m\)

Answer: 3. \(\frac{10}{3} \mathrm{~m}\)

The distance covered in the nth second is

⇒ \(s_n=u+\left(\frac{2 n-1}{2}\right) a\)

Given that u =0,n = 3, a = \(\frac{10}{3}\) ms-2.

∴ The distance covered in the third second is

⇒ \(s_3=0+\left(\frac{2 \times 3-1}{2}\right) \frac{4}{3} \mathrm{~m}=\frac{10}{3} \mathrm{~m}\)

Question 8. A car moves from X to Y with a uniform speed v1 and returns to X with a uniform speed v2. The average speed during this round trip is

  1. \(\sqrt{v_1 v_2}\)
  2. \(\frac{v_1 v_2}{v_1+v_2}\)
  3. \(\frac{v_1+v_2}{2}\)
  4. \(\frac{2 v_1 v_2}{v_1+v_2}\)

Answer: 4. \(\frac{2 v_1 v_2}{v_1+v_2}\)

Average speed = \(\frac{\text { total distance covered }}{\text { total time taken }}\)

⇒ \(\frac{s+s}{\frac{s}{v_1}+\frac{s}{v_2}}=\frac{2 s}{s\left(\frac{1}{v_1}+\frac{1}{v_2}\right)}=\frac{2 v_1 v_2}{v_1+v_2}\)

Question 9. A particle moves along a straight line OX. At a time t (in seconds), the distance x (in meters) of the particle from O is given by \(x=40+12 t-t^3\). How long would the particle travel before coming to rest?

  1. 16 m
  2. 24 m
  3. 40 m
  4. 50 m

Answer: 1. 16 m

Given that x = 40+12t-t3.

∴ instantaneous velocity, v = \(\frac{d x}{d t}\) =12-3t².

For v = 0,

12-3t2 = 0 => t = 2s.

The distance covered by the particle before coining to rest will be

⇒ \(s=\int d s=\int_{t=0}^{t=2 s} v d t=\int_0^{2 s}\left(12-3 t^2\right) d t\)

⇒ \(\left[12 t-t^3\right]_0^2=\left(12 \times 2-2^3\right) \mathrm{m}\)

= 16 m.

Question 10. The position s (in meters) of a particle as a function of time t (in seconds) is expressed as \(s=3 t^3+7 t^2+14 t+8\) The value of the acceleration of the particle at time t =1 s is

  1. 16 ms-2
  2. 32 ms-2
  3. 10 ms-2
  4. 30 ms-2

Answer: 2. 32 ms-2

Given that s = 3³ +7t² +14t +8.

Hence, velocity = \(v=\frac{d s}{d t}=9 t^2+14 t+14\)

and acceleration = \(a=\frac{d v}{d t}=18 t+14\)

The value of the acceleration t =1 s is

⇒ \(|a|_{t=1 \mathrm{~s}}=|18 \mathrm{t}+14|_{t=1 \mathrm{~s}}\)

=32 ms-2.

Question 11. The position x of a particle varies with time t as \(x=a t^2-b t^3\). The time at which acceleration will be zero is equal to

  1. \(\frac{a}{3 b}\)
  2. \(\frac{3 a}{2 b}\)
  3. \(\frac{a}{b}\)
  4. Zero

Answer: 1. \(\frac{a}{3 b}\)

The position (x) of the given particle as a function of time (t) is

⇒ \(x=a t^2-b t^3\)

∴ instantaneous velocity = \(v=\frac{d x}{d t}=2 a t-3 b t^2\)

and instantaneous acceleration = \(a=\frac{d v}{d t}=2 a-6 b t\)

When A = 0, the required time is t = \(\frac{2 a}{6 b}=\frac{a}{3 b}\)

Question 12. The displacement * of a particle varies with the time t as \(x=a e^{-\alpha t}+b e^{\beta t}\), where a, b, a and p are positive constants. The velocity of the particle will

  1. Drop to zero when a = p
  2. Go on decreasing with time
  3. Be independent of p
  4. Go on increasing with time

Answer: 4. Go on increasing with time

Given that \(x=a e^{-\alpha t}+b e^{\beta t}\)

Hence, velocity = \(v=\frac{d x}{d t}=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\)

The velocity of the particle will increase with time, being an exponential function

Question 13. A car is moving along a straight road with a uniform acceleration. It passes two points P and Q separated by some distance with a velocity of 30 km h-1 and 40 km h-1 respectively. The velocity of the car while crossing the midpoint between P and Q will be

  1. 33.5 km h-1
  2. \(20 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)
  3. \(25 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)
  4. 38 km h-1

Answer: 3. \(25 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)

Given that \(v_{\mathrm{P}}=30 \mathrm{~km} \mathrm{~h}^{-1} \text { and } v_{\mathrm{Q}}=40 \mathrm{~km} \mathrm{~h}^{-1}\)

Let the velocity at the midpoint O be v0 and the uniform acceleration
be a.

Motion In A Staight Line multiple choice question and answers velocity at the midpoint Q 13

For the motion from P to Q,

⇒ \(v_{\mathrm{Q}}^2-v_{\mathrm{P}}^2=2 a L\)

or, \(40^2-30^2=2 a L\) ….(1)

For the motion from to Q,

⇒ \(v_{\mathrm{Q}}{ }^2-v_{\mathrm{O}}{ }^2=2 a\left(\frac{L}{2}\right)\)

or, \(40^2-v_{\mathrm{O}}^2=a L .\) ….(2)

Dividing (1) by (2),

⇒ \(\frac{40^2-30^2}{40^2-v_O^2}=2\)

Solving, we get \(v_0=25 \sqrt{2} \mathrm{kmh}^{-1}\).

Question 14. The acceleration of a particle is increasing linearly with time as bt. The particle starts from the origin with an initial velocity v0. The distance traveled by the particle in time t will be

  1. \(v_0 t+\frac{1}{3} b t^2\)
  2. \(v_0 t+\frac{1}{6} b t^3\)
  3. \(v_0 t+\frac{1}{2} b t^2\)
  4. \(v_0 t+\frac{1}{3} b t^3\)

Answer: 2. \(v_0 t+\frac{1}{6} b t^3\)

Acceleration = a = bt = \(\frac{dv}{dt}\).

So,dv = bt dt.

Integrating,

⇒ \(\int^v d v=b \int^t t d t \Rightarrow v-v_0=\frac{b}{2} \cdot t^2 \Rightarrow v=\frac{d x}{d t}=v_0+\frac{b t^2}{2}\)

Integrating again,

⇒ \(x=\int v d t=\int_0^t\left(v_0+\frac{b t^2}{2}\right) d t=\left[v_0 t+\frac{b t^3}{6}\right]_0^t\)

∴ the distance traveled in the time this

⇒ \(x=v_0 t+\frac{b t^3}{6}\)

Question 15. A particle starts sliding down a smooth inclined plane. If sn is the distance traversed by it from a time t = (n-1) s to a time t = n s then the ratio \(s_n / s_{n+1}\) is

  1. \(\frac{2 n-1}{2 n+1}\)
  2. \(\frac{2 n+1}{2 n}\)
  3. \(\frac{2 n}{2 n+1}\)
  4. \(\frac{2 n+1}{2 n-1}\)

Answer: 1. \(\frac{2 n-1}{2 n+1}\)

Given that u = 0. So, the distance covered during the nth second is

⇒ \(s_n=u+(2 n-1) \frac{a}{2}=(2 n-1) \frac{a}{2}\)

and that during the (n + l)th second is

⇒ \(s_{n+1}=[2(n+1)-1] \frac{a}{2}\)

∴ \(\frac{S_n}{S_{n+1}}=\frac{2 n-1}{2 n+1}\)

Question 16. A particle is moving with a velocity \(\vec{v}=k(y \hat{i}+x \hat{j})\), where k is constant. The general equation for its path is

  1. y² – x² + constant
  2. y = x² + constant
  3. y² = x + constant
  4. xy = constant

Answer: 1. y² – x² + constant

Given that \(\vec{v}=k(y \hat{i}+x \hat{j})=v_x \hat{i}+v_y \hat{j}\)

∴ \(v_x=\frac{d x}{d t}=k y \text { and } v_y=\frac{d y}{d t}=k x\)

∴ \(\frac{v_y}{v_x}=\frac{x}{y}=\frac{d y / d t}{d x / d t}=\frac{d y}{d x}\)

∴ ydy = xdx.

Integrating,

⇒ \(\frac{y^2}{2}=\frac{x^2}{2}\) + constant

y² = x² + constant.

Question 17. A particle, located at x = 0 at a time t = 0, starts moving along the positive x-direction with a velocity v, which varies as v = a√x. The displacement of the particle varies with time as

  1. t

Answer: 2. t²

Instantaneous velocity = \(v=\frac{d x}{d t}=\alpha \sqrt{x}\)

∴ \(\frac{d x}{\sqrt{x}}=\alpha d t\)

Integrating,

⇒ \(2 \sqrt{x}=\alpha t \Rightarrow 4 x=\alpha^2 t^2 \Rightarrow x \propto t^2\).

Question 18. The velocity of a particle is \(v=v_0+g t+f t^2\). If its position at t= 0 gt is x = 0 then its displacement after a unit time (f = 1 s) is

  1. \(v_0+2 g+3 f\)
  2. \(v_0+\frac{g}{2}+\frac{f}{3}\)
  3. \(v_0+g+f\)
  4. \(v_0+\frac{g}{2}+f\)

Answer: 2. \(v_0+\frac{g}{2}+\frac{f}{3}\)

Given that \(v=\frac{d x}{d t}=v_0+g t+f^2\).

So, the displacement at a time t will be

⇒ \(x=\int d x=\int v d \dot{t}=\int\left(v_0+g t+f t^2\right) d t\)

⇒ ⇒ \(\left[v_0 t+g \frac{t^2}{2}+f \frac{t^3}{3}\right]_{t=0}^{t=1 \mathrm{~s}}\)

⇒ \(v_0+\frac{g}{2}+\frac{f}{3}\)

Question 19. The relation between the time t and the displacement x is t = ax² + bx, where a and b are constants. The acceleration is

  1. -2abv²
  2. 2bv³
  3. -2av³
  4. 2av²

Answer: 3. -2av³

Given that t = ax² + bx.

Differentiating with respect to time t,

⇒ \((2 a x+b) \frac{d x}{d t}=1\)

or v(2ax + b) =1

or \(v=\frac{1}{2 a x+b}\)

Differentiating again with respect to time t,

acceleration = \(\frac{d v}{d t}=\frac{-2 a v}{(2 a x+b)^2}=-(2 a v) v^2=-2 a v^3\)

Question 20, The speeds of two identical cars are u and 4u at a given instant. The ratio of their respective distances at which the two cars are stopped after that instant is

  1. 1:1
  2. 1:4
  3. 1:8
  4. 1:16

Answer: 4. 1:16

If d = stopping distance then

⇒ \(v^2=0=u^2-2 a d \text { or } d=\frac{u^2}{2 a}\)

∴ \(\frac{s_1}{s_2}=\frac{u_1^2}{u_2^2}\) [∵ a is the same for the two identical cars]

\(\frac{u^2}{(4 u)^2}\)

= 1:16.

Question 21. An object moving with a speed of 6.25 m s-1 is decelerated at a rate given \(\frac{d v}{d t}=-2.5 \sqrt{v}\), where v is the instantaneous speed. The time taken by the object to come to rest would be

  1. 2 s
  2. 4 s
  3. 8 s
  4. 1 s

Answer: 1. 2 s

Given that \(\frac{d v}{d t}=-2.5 \sqrt{v}\)

∴ \(\frac{d v}{\sqrt{v}}=-2.5 d t\)

Integrating, 2√v = -2.5f + c.

At t = 0, v = 6.25ms-1, soc = 5.

For the object to come to rest,

2V0 = -2.5t + 5

=> t = 2s

Question 22. A particle is moving eastwards with a velocity of 5 m s-1. In 10 s the velocity changes to 5 m s-1 northwards. The average acceleration during tins time is

  1. \(\frac{1}{\sqrt{2}} \mathrm{~m} \mathrm{~s}^{-2} \text { towards } \mathrm{NE}\)
  2. \(\frac{1}{2} \mathrm{~m} \mathrm{~s}^{-2} \text { towards north }\)
  3. Zero
  4. \(\frac{1}{\sqrt{2}} \mathrm{~m} \mathrm{~s}^{-2} \text { towards NW }\)

Answer: 4. \(\frac{1}{\sqrt{2}} \mathrm{~m} \mathrm{~s}^{-2} \text { towards NW }\)

The average acceleration is

⇒ \(\vec{a}_{\mathrm{av}}=\frac{\text { change in velocity }}{\text { time }}=\frac{\vec{v}_{\mathrm{f}}-\vec{v}_{\mathrm{i}}}{t}\)

⇒ \(\frac{\left(5 \mathrm{~ms}^{-1}\right) \hat{j}-\left(5 \mathrm{~ms}^{-1}\right) \hat{i}}{10 \mathrm{~s}}=\frac{1}{2}(\hat{j}-\hat{i}) \mathrm{ms}^{-2}\)

∴ \(\left|\vec{a}_{\mathrm{av}}\right|=\frac{\sqrt{2}}{2} \mathrm{~ms}^{-2}=\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2} \text { (towards NW)}\)

Question 23. The adjoining figure shows the variation of the acceleration of a particle with time. The particle is moving along a straight line and has a velocity v = 2 m s-1 at f = 0. The velocity after 2 s will be

Motion In A Staight Line multiple choice question and answers variation of the acceleration Q 23

  1. 2 ms-1
  2. 4 ms-1
  3. 6 ms-1
  4. 8 m s-1

Answer: 3. 6 ms-1

The area enclosed under the given-t graph is

⇒ \(A=\frac{1}{2}(2 \mathrm{~s})\left(4 \mathrm{~ms}^{-2}\right)=4 \mathrm{~ms}^{-1}\)

But this area = \(\int a d t=\int \frac{d v}{d t} d t=[v]_{v_{\mathrm{i}}}^{v_{\mathrm{i}}}\)

∴ 4 m s-1 = vf – vi = vf-2 m s-1

∴ The velocity after 2 s is vf = 6 m s-1.

Question 24. The given figure shows the velocity-time graph of two particles A and B moving along the same straight line in the same direction. Which of the following statements is true?

Motion In A Staight Line multiple choice question and answers variation of the acceleration Q 23

  1. The relative velocity is zero.
  2. The relative velocity is nonzero but constant.
  3. The relative velocity increases continuously.
  4. The relative velocity decreases continuously.

Answer: 3. The relative velocity increases continuously.

The relative velocity (ur) and relative acceleration (ar) of two particles

are related by vr = ur + art.

Here, ur = 0 and at = (aA – aB) = constant.

∴ relative velocity = vT = (aA- aB)t.

∴ \(v_{\mathrm{r}} \propto t .\)

Hence, the relative velocity increases continuously with time.

Question 25. The position of a particle moving along a straight line is expressed as a function of time (t) as x = 6 + 12t- 2t², where x is in meters and t is in seconds. The distance covered by the particle in the first 5 s is

  1. 32 m
  2. 24 m
  3. 20 m
  4. 26 m

Answer: 4. 26 m

Given that x = 6 + 12t- 2t².

Hence, velocity = u = \(v=\frac{d x}{d t}=12-4 t\)

and acceleration = \(a=\frac{d v}{d t}=-4 \mathrm{~m} \mathrm{~s}^{-2}\)

Initial velocity = \(u=[12-4 t]_{t=0}=12 \mathrm{~m} \mathrm{~s}^{-1}\)

Due to the retardation of -4 ms-2, the particle comes to rest when
u = 0 =12-4t or t =3s.

The distance covered in3s is

⇒ \(s_1=u t+\frac{1}{2} a t^2=(12 \times 3) \mathrm{m}+\frac{1}{2}(-4)(9) \mathrm{m}=18 \mathrm{~m}\)

The distance covered backward in 2s is

⇒ \(s_2=\frac{1}{2}(4)(2)^2 \mathrm{~m}=8 \mathrm{~m}\)

Hence, total distance = s = s1 + s2

= 18 m + 8 m

= 26 m.

Question 26. A particle starts moving along a straightlinesuchthattheacceleration varies with the displacement (s) as shown in the given figure. Which of the following represents the velocity-displacement graph?

Motion In A Staight Line multiple choice question and answers velocity displacement graph Q 26

Motion In A Staight Line multiple choice question and answers velocoty displacement graphsQ 26

Answer: 2.

Motion In A Staight Line multiple choice question and answers velocity displacement graph Q 26.

The graph showing the variation of acceleration (a) with displacement (s) is linear. Hence, a = ks, where k- constant,

∴ \(a=\frac{d v}{d t}=\frac{d v}{d s} \frac{d s}{d t}=k s\)

or, \(v \frac{d v}{d s}=k s\) [∵ \(\frac{d s}{d t}=v\)]

or, vdv = ks ds.

Integrating,

⇒ \(\frac{v^2}{2}=\frac{k s^2}{2} \text { or } v \propto s\).

⇒ Question 27. Two particles moving along the same straight line and leaving the same point O at the same time have their initial velocities u and 2u and uniform accelerations 2a and a respectively. The distance of the two particles from O when one particle overtakes the other is

  1. \(\frac{u^2}{a}\)
  2. \(\frac{4 u^2}{a}\)
  3. \(\frac{6 u^2}{a}\)
  4. \(\frac{8 u^2}{a}\)

Answer: 3. \(\frac{6 u^2}{a}\)

When one particle overtakes the other at a time t, both travel the same distance s.

Hence, for the first particle,

⇒ \(s=u t+\frac{1}{2}(2 a) t^2\)

and for the second particle,

⇒ \(s=2 u t+\frac{1}{2} a t^2\)

Equating,

⇒ \(u t+a t^2=2 u t+\frac{a}{2} t^2 \Rightarrow \frac{a}{2} t=u \Rightarrow t=\frac{2 u}{a} .\)

Hence,

⇒ \(s=u t+\frac{1}{2}(2 a) t^2=u\left(\frac{2 u}{a}\right)+\frac{1}{2}(2 a)\left(\frac{2 u}{a}\right)^2=\frac{2 u^2}{a}+\frac{4 u^2}{a}=\frac{6 u^2}{a}\)

Question 28. A particle moves a distance x in a time t according to the equation x = {t + 5)-1. The acceleration of the particle is proportional to

  1. (velocity)2/3
  2. (distance)²
  3. (distance)-2
  4. (velocity)3/2

Answer: 1. (distance)2

The displacement (x) of the given particle at a time t is x = (t + 5)-1.

∴ velocity = \(v=\frac{d x}{d t}=-(t+5)^{-2}\)

and acceleration = \(a=\frac{d v}{d t}=2(t+5)^{-3}\)

∴ \(a=2\left[(t+5)^{-1}\right]^3=2 x^3 \Rightarrow a \propto x^3\)

But

⇒ \(v^{3 / 2}=-(t+5)^{-3} \text {. Hence, } a \propto v^{3 / 2} \text {. }\)

Question 29. A particle starts moving from rest under the action of a constant force. If it covers a distance of sx in the first s and a distance s² in the first 20 s then

  1. s2 = s1
  2. s2 = 4s1
  3. s2 = 3s1
  4. s2 = 2s1

Answer: 2. s2 = 4s1

Initial velocity = u = 0.

The distance covered in the first s is

⇒ \(s_1=\frac{1}{2} a\left(10^2\right)=50 a,\)

and the distance covered in the first 20 s is

⇒ \(s_2=\frac{1}{2} a\left(20^2\right)=200 a=4(50 a)=4 s_1\)

∴ \(s_2=4 s_1\)

Question 30. A particle moves in a straight line with a constant acceleration. It changes its velocity from 10ms-1 to 20ms-1 while passing through a distance of 135 m in a time t. The value of t is

  1. 10 s
  2. 9 s
  3. 12 s
  4. 1.8 s

Answer: 2. 9 s

Given that, u = 10 m s-1, v = 20m s-1, and s = 135 m.

∴ acceleration = \(a=\frac{v^2-u^2}{2 s}=\frac{400-100}{2 \times 135} \mathrm{~ms}^{-2}=\frac{10}{9} \mathrm{~ms}^{-2} .\)

Now, v = u + at, so the required time is

⇒ \(t=\frac{v-u}{a}=\frac{20 \mathrm{~ms}^{-1}-10 \mathrm{~ms}^{-1}}{\frac{10}{9} \mathrm{~ms}^{-2}}=9 \mathrm{~s}\)

Question 31. The position x of a particle with respect to the time t along the x-axis is given by \(x=9 t^2-t^3\), where x is in meters and t is in seconds. What will be the position of the particle when it achieves the maximum speed in the positive

  1. 54 m
  2. 32m
  3. 24 m
  4. 81m

Answer: 1. 54 m

The position as a function of time (t) is x = 9t²-t³ and thus the velocity

is \(v=\frac{d x}{d t}=18 t-3 t^2\)

For v to be the maximum,

\(\frac{d v}{d t}\) = 0 or 18 — 6f = 0 or t = 3s.

The position at f = 3 s willbe

⇒ \(|x|_{t=3 \mathrm{~s}}=\left|9 t^2-t^3\right|_{t=3 \mathrm{~s}}\)

= 81m-27m

= 54m.

Question 32. A bus is moving at a speed of 10 ms-1 on a straight road. A scooterist wishes to overtake the bus in 100s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?

  1. 20 ms-1
  2. 10 ms-1
  3. 40 ms-1
  4. 25 ms-1

Answer: 1. 20 ms-1

Let the velocity of the scooter be v sc.

Given that the velocity of the bus = 10 m s-1 and its initial (relative) separation between the two vehicles =1 km = 1000 m.

The acceleration values of both are zero. So, for the relative separation to reduce to zero,

⇒ \(s_{\mathrm{r}}=u_{\mathrm{r}} t \Rightarrow t=\frac{s_{\mathrm{r}}}{u_{\mathrm{r}}}=\frac{1000 \mathrm{~m}}{v_{\mathrm{sc}}-10 \mathrm{~m} \mathrm{~s}^{-1}}=100 \mathrm{~s}\)

⇒ \(v_{\mathrm{sc}}-10 \mathrm{~m} \mathrm{~s}^{-1}=10 \mathrm{~m} \mathrm{~s}^{-1} \Rightarrow v_{\mathrm{sc}}=20 \mathrm{~m} \mathrm{~s}^{-1}\)

Hence, the velocity of the scooter is v sc = 20 m s-1.

Question 33. If a car at rest accelerates uniformly to a speed of 144 km h-1 in 20 s, it covers a distance of

  1. 2980 m
  2. 1440 m
  3. 400 m
  4. 20 m

Answer: 3. 400 m

Given that u = 0, v = 144 km h-1  = 144 x \(\frac{5}{18}\) m s-1  = 40 m s-1  and the time taken = 20 s.

∵ v = u + at,

∴ \(a=\frac{v}{t}=\frac{40 \mathrm{~m} \mathrm{~s}^{-1}}{20 \mathrm{~s}}=2 \mathrm{~m} \mathrm{~s}^{-2}\)

∴ The distance covered is

⇒ \(s=u t+\frac{1}{2} a t^2=0+\frac{1}{2}\left(2 \mathrm{~m} \mathrm{~s}^{-2}\right)(20 \mathrm{~s})^2\)

= 400m.

Question 34. A car accelerates from rest at a constant rate for some time, after which it decelerates at a constant rate of P and comes to rest. If the total time elapsed is t, the maximum velocity acquired by the car will be

  1. \(\left(\frac{\alpha^2+\beta^2}{\alpha \beta}\right) t\)
  2. \(\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\)
  3. \(\left(\frac{\alpha^2-\beta^2}{\alpha \beta}\right) t\)
  4. \(\left(\frac{\alpha+\beta}{\alpha \beta}\right) t\)

Answer: 2. \(\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\)

Let v max be the maximum velocity. While accelerating, sa

⇒ \(v_{\max }=\alpha t_1 \Rightarrow t_1=\frac{v_{\max }}{\alpha} .\)

Similarly, while decelerating to stop finally,

⇒ \(\frac{0-v_{\max }}{t_2}=-\beta \text { or } t_2=\frac{v_{\max }}{\beta}\)

Total time = \(t=t_1+t_2=v_{\max }\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)

∴ \(v_{\max }=\left(\frac{\alpha \beta}{\alpha+\beta}\right)\)

Question 35. 11M? displacement 9 (in meters) of a particle moving along a straight line is expressed as a function of time t (in seconds) by the equation \(s=\left(t^3-6 t^2+3 t+4\right)\). When the acceleration is zero, the velocity will be

  1. 42 m s-1
  2. -9 m s-1
  3. 5ms-1
  4. -12 m s-1

Answer: 2. -9 m s-1

Given that s = t³-6t² + 3t + 4.

Hence, velocity = \(y=\frac{d s}{d t}=3 t^2-12 t+3\)

and acceleration = a = \(\frac{d v}{d t}\)

= 6t-12 = 0.

So, t = 2 s.

The required velocity is

⇒ \(v=\left|3 t^2-12 t+3\right|_{t=2 \mathrm{~s}}\)

= (12-24+3) ms-1

= -9 ms-1.

Question 36. A bus travels the first one-third distance at a speed of 10 km h-1 the next one-third distance at 20 km h-1 and the last one-third distance at 60 km h-1. The average speed of the bus is

  1. 10 km h-1
  2. 16 km h-1
  3. 18 km h-1
  4. 40 km hr-1

Answer: 3. 18 km h-1

Let the total distance covered by s.

∴ \(T=t_1+t_2+t_3=\frac{s}{3 u_1}+\frac{s}{3 u_2}+\frac{s}{3 u_3}=\frac{s}{3}\left(\frac{1}{u_1}+\frac{1}{u_2}+\frac{1}{u_3}\right)\)

Hence, the total time of travel is

⇒ \(\frac{s}{3}\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{60}\right)=\frac{s \times 10}{3 \times 60}\)

∴ average speed \(=v_{\mathrm{av}}=\frac{\mathrm{s}}{T}=18 \mathrm{~km} \mathrm{~h}^{-1}\)

Question 37. A car covers a distance of 200 m. It covers the first half of the distance at a speed of 40 km h-1 and the second half at a speed v. If the average speed of the car is 48 km h-1, the value of vis

  1. 48 km h-1
  2. 45 km h-1
  3. 60 km h-1
  4. 50 km h-1

Answer: 3. 60 km h-1

Since the motion of the car is uniform, its acceleration is zero. During the first half,

⇒ \(\frac{s}{2}=u_1 t_1 \quad \text { or } \quad t_1=\frac{s}{2 u_1}\)

and during the second half,

⇒ \(t_2=\frac{s}{2 u_2}\)

the total time of travel is

⇒ \(T=t_1+t_2=\frac{s}{2}\left(\frac{1}{u_1}+\frac{1}{u_2}\right)\)

the average speed is

⇒ \(v_{\mathrm{av}}=\frac{s}{T}=\frac{2}{\frac{1}{u_1}+\frac{1}{u_2}}\)

Substituting the values,

⇒ \(48=\frac{2}{\frac{1}{40}+\frac{1}{v}}\)

or, \(\frac{1}{40}+\frac{1}{v}=\frac{1}{24}\)

or, v = 60 km-1

Question 38. A car starting from rest and moving with a constant acceleration covers a distance of s1 in the fourth second and a distance of s2 in the sixth second The ratio s1/s2 is

  1. \(\frac{4}{9}\)
  2. \(\frac{6}{11}\)
  3. \(\frac{7}{11}\)
  4. \(\frac{2}{3}\)

Answer: 3. \(\frac{7}{11}\)

Given that u = 0. So, in the fourth second,

⇒ \(s_4=u+\left(\frac{2 n-1}{2}\right) a=\frac{7}{2} a\)

and in the sixth second

⇒ \(s_6=\frac{11}{2} a .\)

∴ \(\frac{s_4}{s_6}=\frac{\frac{7}{2} a}{\frac{11}{2} a}=\frac{7}{11}\)

Question 39. The acceleration a of a body starting from rest varies with the time t according to the relation a = 3t + 4. The velocity of the body at the time t = 2s will be

  1. 10 ms-1
  2. 12 ms-1
  3. 14 m s-1
  4. 16 m s-1

Answer: 3. 14 m s-1

Given that

⇒ \(a=\frac{d v}{d t}=3 t+4\)

∴ \(v=\int_0^t(3 t+4) d t=\frac{3}{2} t^2+4 t\)

Aat t=2s, \(v=\left[\frac{3}{2}\left(2^2\right)+4(2)\right] \mathrm{ms}^{-1}\)

= 143s -1.

Question 40. The displacement y meters) of a body varies with the time t Cm seconds) as \(y=-\frac{2}{3} t^2+16 t+2\) How long does the body take to come to rest? 

  1. 8 s
  2. 10 s
  3. 12s
  4. 16 s

Answer: 3. 12s

The position is given by

⇒ \(y=-\frac{2}{3} t^2+16 t+2\)

Comparing the equation with the standard equation of a uniformly accelerated motion,

⇒ \(s=u t+\frac{1}{2} a t^2\) we have

u = 16ms-1

and \(\frac{a}{2}=-\frac{2}{3} \text { or } a=-\frac{4}{3} \mathrm{~ms}^{-2}\)

For the body to come to rest, v = 0.

∴ \(v=u+a t \Rightarrow 0=16 \mathrm{~m} \mathrm{~s}^{-1}+\left(-\frac{4}{3} \mathrm{~m} \mathrm{~s}^{-2}\right) t\)

t = 12s.

Question 41. The given figure represents the speed-time graph of a body moving along a straight line. How much distance does it cover during the last 10 seconds of its motion?

Motion In A Staight Line multiple choice question and answers speed time graph Q 41

  1. 40m
  2. 100 m
  3. 60 m
  4. 120 m

Answer: 3. 60 m

The area under the v-t graph gives the displacement. So, the distance covered in the last 10 s will be the area of ΔABC, i.e.,

⇒ \(\frac{1}{2}(B C)(A B)\)

⇒ \(\frac{1}{2}(20 \mathrm{~s}-10 \mathrm{~s})\left(20 \mathrm{~m} \mathrm{~s}^{-1}\right)\)

= 100m.

Question 42. The velocity-time graph of a stone thrown vertically upwards with an initial velocity of 30 m s-1 is shown in the given figure. The velocity in the upward direction is taken as positive. What is the maximum height to which the stone rises?

Motion In A Staight Line multiple choice question and answers velocity time graph Q 42

  1. 30m
  2. 45m
  3. 80m
  4. 90m

Answer: 2. 45m

Given that u = 30m s-1.

Acceleration = slope of the line AB, so a = \(\frac{0-30 \mathrm{~ms}^{-1}}{3 \mathrm{~s}}=-10 \mathrm{~m} \mathrm{~s}^{-2}\)

At the maximum height, v = 0.

Hence, by \(v^2=u^2-2 g h\)

⇒ \(h=\frac{u^2}{2 g}=\frac{900}{20} \mathrm{~m}\)

= 45m

Question 43. The driver of train A moving at a speed of 30 m s-1 sights another train B moving on the same track at a speed of 10 m s-1 in the same direction. He immediately applies the brakes and achieves a uniform retardation of 2 m s-2. To avoid a collision, what must be the minimum distance between the two trains?

  1. 100 m
  2. 120 m
  3. 60m
  4. 160 m

Answer: 1. 100 m

The relative speed of the train A with respect to the train B,

vAB = vA -vB

= (30- 10) m s-1

= 20 m s-1.

To avoid a collision, let the minimum separation between them be s.

This means that the relative speed must reduce to zero when the distance covered is s.

Thus,

⇒ \(v_{\mathrm{r}}^2-u_{\mathrm{r}}^2=2 a_{\mathrm{r}} s\)

⇒ \(0-\left(20 \mathrm{~m} \mathrm{~s}^{-1}\right)^2=2\left(-2 \mathrm{~m} \mathrm{~s}^{-2}\right) \mathrm{s}\)

s = 100m.

Question 44. Three particles P, Q, and R are projected from the top of a tower with the same speed u. P is thrown straight up, Q is thrown straight down and R is thrown horizontally. They hit the ground with speeds of vp, vQ, and vR respectively. Then,

  1. \(v_{\mathrm{P}}>v_{\mathrm{Q}}=v_{\mathrm{R}}\)
  2. \(v_{\mathrm{P}}=v_{\mathrm{Q}}>v_{\mathrm{R}}\)
  3. \(v_{\mathrm{P}}=v_{\mathrm{Q}}=v_{\mathrm{R}}\)
  4. \(v_{\mathrm{P}}>v_{\mathrm{Q}}>v_{\mathrm{R}}\)

Answer: 3. \(v_{\mathrm{P}}=v_{\mathrm{Q}}>v_{\mathrm{R}}\)

The net vertical displacement for each of the three particles is the same (= h).

For P, work done by gravity is mph.

Hence, change in KE = \(\frac{1}{2} m_{\mathrm{P}}\left(v_{\mathrm{P}}^2-u^2\right) \Rightarrow \frac{1}{2} m_{\mathrm{P}}\left(v_{\mathrm{P}}^2-u^2\right)=m_{\mathrm{P}} g h\)

vp² =2gh+u².

Similarly, \(v_{\mathrm{O}}^2=2 g h+u^2 \text { and } v_{\mathrm{R}}^2=2 g h+u^2\)

∴ \(v_{\mathrm{P}}=v_{\mathrm{Q}}=v_{\mathrm{R}}\)

Question 45. A body dropped from a tower of height h covers a distance h/2inthe last second of its motion. The height of the tower is approximately (taking g = 10 ms-1)

  1. 50 m
  2. 55 m
  3. 58 m
  4. 60 m

Answer: 3. 58 m

Let the total time of fall be t. So,

⇒ \(h=\frac{1}{2} g t^2\) ……(1)

The distance covered in the final second is h/2. So, for the upper half (h/2), time = t-1.

∴ \(\frac{h}{2}=\frac{1}{2} g(t-1)^2\) …..(2)

Now, dividing (1) by (2),

⇒ \(2=\left(\frac{t}{t-1}\right)^2\)

Solving, we get

⇒ \(t=(2+\sqrt{2}) \mathrm{s}\)

the height of the tower is

⇒ \(h=\frac{1}{2}(10)(2+\sqrt{2})^2 m=58.3 m \approx 58 m\)

Question 46. A body is projected vertically upwards with a velocity u. It crosses three points A, B, and C in succession in its upward journey with velocities u/2, u/3, and u/4. The ratio AB/BC is

  1. \(\frac{10}{7}\)
  2. \(\frac{20}{7}\)
  3. \(\frac{3}{2}\)
  4. \(\frac{2}{1}\)

Answer: 2. \(\frac{20}{7}\)

For the motion along AB,

⇒ \(\left(\frac{u}{3}\right)^2-\left(\frac{u}{2}\right)^2=-2 g(A B) \Rightarrow A B=\frac{5 u^2}{72 g}\)

Similarly, for the motion along BC,

⇒ \(\left(\frac{u}{4}\right)^2-\left(\frac{u}{3}\right)^2=-2 g(B C)^{\prime} \Rightarrow B C=\frac{7 u^2}{288 g}\)

Hence,

⇒ \(\frac{A B}{B C}=\frac{5}{72} \times \frac{288}{7}=\frac{20}{7}\)

Motion In A Staight Line multiple choice question and answers Q 46

Question 47. From the top of a 40-m-high tower, a stone is projected vertically upwards with an initial velocity of 10 m s-1. After how much time will the stone hit the ground? (Take g = 10 m s-2.)

  1. 4 s
  2. 1 s
  3. 2 s
  4. 3 s

Answer: 1. 4 s

Let us take the point of rejection as the origin. Applying the coordinate sign convention, velocity of projection = u = +10 ms-1 net displacement

= h = -40 m and g =-10 m s-2.

Thus, \(-h=+u t-\frac{1}{2} g t^2\)

or, -40 = 10t -5t²

or t²-2t-8=0 or (t-4)(t + 2)=0.

The acceptable solution is = 4s.

Question 48. In the preceding question, what will be the speed of the stone when it hits the ground?

  1. 20 ms-1
  2. 30 m s-1
  3. 35m sr-1
  4. 40 m s-1

Answer: 3. 35m srl

Applying the standard equation v = u+gt, we have

-v =10 m s-1-10 m s-2 x 4s

= -30 m s-1

v = 30ms-1.

Question 49. A stone dropped from the top of a tower hits the ground after 4 s. How much time does it take to travel the first half of the distance from the top of the tower?

  1. 1 s
  2. 2√2 s
  3. 2 s
  4. √3 s

Answer: 2. 2√2 s

For the total flight of the stone,

⇒ \(h=\frac{1}{2} g t^2\) [∵ u=0]

⇒ \(h=\frac{1}{2} \times 10 \times 4^2=80\)

⇒ \(\frac{h}{2}=40=\frac{1}{2} \cdot(10) t^2=5 t^2\)

So, the required time

⇒ \(t=2 \sqrt{2} \mathrm{~s}\)

Question 50. A stone is projected vertically downwards with a velocity u from the top of a tower. It strikes the ground with a velocity of 3M. The time taken by the stone to reach the ground is

  1. \(\frac{u}{g}\)
  2. \(\frac{3 u}{g}\)
  3. \(\frac{2 u}{g}\)
  4. \(\frac{4 u}{g}\)

Answer: 3. \(\frac{2 u}{g}\)

For the downward motion,

v = u+gt

⇒ 3u = u +gt

⇒ \(t=\frac{2 u}{g}\)

Question 51. In the preceding question, what is the height of the tower?

  1. \(\frac{u^2}{g}\)
  2. \(\frac{2 u^2}{g}\)
  3. \(\frac{3 u^2}{8}\)
  4. \(\frac{4 u^2}{g}\)

Answer: 4. \(\frac{4 u^2}{g}\)

Let h be the height of the tower.

Hence,

⇒ \(h=u t+\frac{1}{2} g t^2=u\left(\frac{2 u}{g}\right)=\frac{1}{2} g\left(\frac{2 u}{g}\right)^2=\frac{4 u^2}{g}\)

Question 52. A stone falls freely under gravity. It covers the distances hv h2 and h3 in the first 5 s, the second 5 s, and the third 5 s respectively. The relation between h1, h2 and h3 is given by

  1. \(h_2=3 h_1 \text { and } h_3=2 h_2\)
  2. \(h_1=h_2=h_3\)
  3. \(h_1=2 h_2=3 h_3\)
  4. \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

Answer: 4. \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

The stone is dropped from O.

So, \(O A=h_1=\frac{1}{2} g\left(5^2\right)=\frac{25 g}{2}\)

⇒ \(O B=h_1+h_2=\frac{1}{2} g\left(10^2\right)=\frac{100 g}{2}\)

and \(O C=h_1+h_2+h_3=\frac{1}{2} g\left(15^2\right)=\frac{225 g}{2}\)

∴ \(h_2 \doteq O B-O A=\frac{75 g}{2} \text { and }\)

⇒ \(h_3=O C-O B=\frac{125 g}{2}\)

Hence,

⇒ \(h_1: h_2: h_3=25: 75: 125=1: 3: 5\)

∴ \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

Motion In A Staight Line multiple choice question and answers Q 52

Question 53. A ball is dropped from a high-rise platform at t = 0. After 6 s, another ball is thrown downwards from the same platform with a speed v. The two balls meet at t – 18 s. What is the value of 7 (Take g = 10 m s-2.)

  1. 40 ms-1
  2. 75ms-1
  3. 60 ms-1
  4. 55 ms-1

Answer: 2. 75ms-1

The two balls meet at time t=18s. Hence, for the first ball,

⇒ \(h=\frac{1}{2} g t^2=\frac{1}{2}\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(18 \mathrm{~s})^2\)

= 5 (18)² m.

For the second ball, time = (t-6) s when the two balls meet.

Hence,

h = v(t-6)+ \(\frac{1}{2}\)g(t- 6)2

= v(t8- 6) + 5(18- 6)²

=12v + 5 x 12².

Equating the two expressions for h, we have

12v + 5 x 144 = 5 x 18 x 18

⇒ \(v=\frac{5 \times 18^2}{12} \mathrm{~ms}^{-1}-\frac{5 \times 144}{12} \mathrm{~ms}^{-1}=75 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 54. Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from the heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is

  1. \(\frac{12}{5}\)
  2. \(\frac{5}{4}\)
  3. \(\frac{5}{12}\)
  4. \(\frac{4}{5}\)

Answer: 3. \(\frac{5}{12}\)

Theoretically, the time of fall is independent of mass.

So, applying \(h=\frac{1}{2} g t^2\)

or, \(t=\sqrt{\frac{2 h}{g}}\), we obtain

⇒ \(\frac{t_{\mathrm{A}}}{t_{\mathrm{B}}}=\sqrt{\frac{h_{\mathrm{A}}}{h_{\mathrm{B}}}}=\sqrt{\frac{16 \mathrm{~m}}{25 \mathrm{~m}}}=\frac{4}{5}\)

Question 55. A ball is projected vertically upwards. It attains a speed of 10 m s-1 when it reaches half its maximum height. How high does the ball rise? (Take g = 10 m s-2.)

  1. 5 m
  2. 20 m
  3. 10 m
  4. 15 m

Answer: 3. 10 m

v² = u² – 2gh, where o = 0 at the maximum height (h) and u = 10 m s-1.

∴ \(0=10^2-2 g\left(\frac{h}{2}\right)=100-10 h\)

Hence, h = 10 m.

Question 56. If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

  1. \(\frac{1}{2} g t^2\)
  2. ut
  3. \(u t-\frac{1}{2} g t^2\)
  4. (u+gt)t

Answer: 1. \(\frac{1}{2} g t^2\)

Let A be the point at the maximum height, where the speed is reduced to zero. Let AB be the distance during the last f seconds. Hence, time of ascent = time of descent, where BA = AB = h

∴ \(h=u_{\mathrm{A}} t+\frac{1}{2} g t^2=0+\frac{1}{2} g t^2=\frac{1}{2} g t^2\)

Note This value of h is independent of u.

Motion In A Staight Line multiple choice question and answers point at the maximum height Q 56

Question 57. A particle is thrown vertically upwards. Its velocity at half the maximum height is 10 m s-1. The maximum height attained is (taking g = 10 m s-2)

  1. 8 m
  2. 10 m
  3. 16 m
  4. 20 m

Answer: 2. 10 m

During the descent of the particle from the maximum height, the speed will be the same (= 10 m s-1) at half the height. Hence,

⇒ \(v^2=u^2+2 g\left(\frac{h}{2}\right)(\text { from the top) }\)

⇒ \(\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2=0^2+2\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right) \frac{h}{2}\)

⇒ \(h=\frac{100}{10} \mathrm{~m}=10 \mathrm{~m}\)

Question 58. A man throws balls with the same speed vertically upwards, one after another, at an interval of 2 s. What should be the speed of projection so that more than two balls are in the sky at any time? (Given that g = 9.8m s-2.)

  1. More than 19.6 m s-1
  2. At least 9.8 m s-1
  3. Any speed less than 19.6 m s-1
  4. 19.6 m s-1 only

Answer: 1. More than 19.6 m s-1

The time interval between the two balls thrown is 2 s. If more than two balls (or at least three balls) remain in the air, the time of flight of the first ball must be greater than 2 x 2 s = 4 s.

∴ \(T>4 \mathrm{~s} \Rightarrow \frac{2 u}{g}>4 \mathrm{~s} \Rightarrow u>2 g=19.6 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 59. A rubber ball is dropped from a height of 5 m on the ground. On bouncing, it rises to 1.8 m. The fractional loss in the velocity bouncing is (taking g – 10 m s-2)

  1. \(\frac{9}{25}\)
  2. \(\frac{16}{25}\)
  3. \(\frac{2}{5}\)
  4. \(\frac{3}{5}\)

Answer: 3. \(\frac{2}{5}\)

The velocity of the ball when it hits the ground for the first time is

⇒ \(v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 5} \mathrm{~m} \mathrm{~s}^{-1}=10 \mathrm{~m} \mathrm{~s}^{-1}\)

The velocity of rebmmd is \(v^{\prime}=\sqrt{2 g \times 1.8} \mathrm{~m} \mathrm{~s}^{-1}=6 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ the fractional loss in velocity is

⇒ \(\frac{\Delta v}{v}=\frac{v-v^{\prime}}{v}=\frac{10-6}{10}=\frac{2}{5}\)

Question 60. A body dropped from a height h strikes the ground with a velocity of 3 m s-1. Another body of the same mass is projected from the same height h with an initial velocity of 4 m s-1. The final velocity of the second body with which it strikes the ground is (assuming g = 10 m s-2)

  1. 3 ms-1
  2. 5ms-1
  3. 4 ms-1
  4. 12 m s-1

Answer: 2. 5ms-1

For the first body,

⇒ \(v^2=u^2+2 g h\)

or (3 ms-1)² = 0² + 2gh = 2gh.

For the second body,

⇒ \(v^2=\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)^2+2 g h\)

or, \(2 g h=v^2-\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

Equating (1) and (2),

v²-(4ms-1)² = (3ms-1

v = 5m s-1.

Question 61. Water drops fall at regular intervals from a tap 5 m above the ground. The third drop leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g = 10 m s-2.)

  1. 1.25 m
  2. 4.00 m
  3. 3.75 m
  4. 2.50 m

Answer: 3. 3.75 m

Given that h = 5 m. For the time of flight t of the first water drop,

⇒ \(5 \mathrm{~m}=\frac{1}{2} g t^2 \Rightarrow t^2=1 \mathrm{~s}^2\) t = 1 s. Hence, the third drop leaves one second after the first drop.

So, the time gap between two successive drops is (1/2) s.

Now, the distance covered by the second drop after (1/2) s will be

⇒ \(\frac{1}{2} g t^2=5\left(\frac{1}{2}\right)^2 \mathrm{~s}=\frac{5}{4} \mathrm{~s}\)

Thus, the distance of the second drop above the ground will be

⇒ \(5 \mathrm{~m}-\frac{5}{4} \mathrm{~m}=\frac{15}{4} \mathrm{~m}\)

= 3.75 m.

Question 62. A body dropped from the top of a tower fell through 40 m during the last two seconds of its fall. The height of the tower is (assuming g = 10 m s-2)

  1. 45 m
  2. 50 m
  3. 80 m
  4. 60 m

Answer: 1. 45 m

Let h = height of the tower and t = time of flight.

Now, \(h=\frac{1}{2} g t^2 \text { and } h-40=\frac{1}{2} g(t-4)^2\)

Subtracting,

⇒ \(40=\frac{g}{2}\left[t^2-(t-4)^2\right]\)

= 5(2t-4)(4)

=> 2t = 6s

=> t = 3s.

∴ \(h=\frac{1}{2} g t^2=5\left(3^2\right) \mathrm{m}\)

= 45 m

Question 63. What will be the ratio of the distances covered by a freely falling body from rest in the fourth and fifth seconds of its fall?

  1. 1:1
  2. 7:9
  3. 16:25
  4. 4:5

Answer: 2. 7:9

The distance covered in the nth second is given by

⇒ \(s_n=u+\left(\frac{2 n-1}{2}\right) a\)

Hence, for the fourth second,

⇒ \(s_4=\frac{7 g}{2}\)

and for the fifth second,

⇒ \(s_5=\frac{9 g}{2}\) [∵ u = 0]

∴ \(\frac{s_4}{s_5}=\frac{7}{9} \quad \text { or } \quad s_4: s_5=7: 9\)

Question 64. A balloon is rising vertically upwards with a velocity of 10 m s-1, When it is at a height of 45 m above the ground, a parachutist bails out from it. After 3 s, he opens the parachute and decelerates at a constant rate of Sms-2. What was the height of the parachute above the ground when he opened the parachute? (Take g = 10 m s-2.)

  1. 30 m
  2. 15 m
  3. 60 m
  4. 45 m

Answer: 1. 30 m

When the parachutist bails out, he has the velocity of the balloon, which is 10 m s-1 upwards.

∴ the net displacement in 3 s is

⇒ \(s=u t+\frac{1}{2}\left(-g t^2\right)=10 \times 3-\frac{10}{2}(3)^2\)

= -15 m.

Here the negative sign indicates a downward displacement.

Hence, the height above the ground when he opens the parachute is

= 45m-15m

= 30m.

Question 65. In the previous problem, how far was the parachutist from the balloon at t-3 s?

  1. 15 m
  2. 30 m
  3. 45 m
  4. 60 m

Answer: 3. 45 m

The balloon rises up in 3 s by a height of (10 m s-1)(3 s) = 30 m. Hence, the parachutist is now at a distance of 30 m + 15 m = 45 m from the balloon.

Question 66. In the preceding problem, how much time does the parachutist take to hit the ground after his exit from the balloon?

  1. 4 s
  2. 5 s
  3. 6 s
  4. 7 s

Answer: 2. 5 s

The total time the parachutist takes after his exit from the balloon to hit the ground is 3 s + 2 s = 5 s.

Question 67. From the top of a tower, a stone is thrown up and it reaches the ground in a time tx. A second stone is thrown down with the same speed and it reaches the ground in a time t2. A third stone is released from rest and it reaches the ground in a time t3. The relationship of t, t2, and t3 is given by

  1. \(\frac{1}{t_3}=\frac{1}{t_1}+\frac{1}{t_2}\)
  2. \(t_3=\sqrt{t_1 t_2}\)
  3. \(t_3=\frac{1}{2}\left(t_1+t_2\right)\)
  4. \(t_3{ }^2=t_1{ }^2-t_2{ }^2\)

Answer: 2. \(t_3=\sqrt{t_1 t_2}\)

Taking the coordinate sign convention into consideration, we have

for the firststone, \(-h=u t_1-\frac{1}{2} g t_1^2\) …(1)

for the second stone, \(h=u t_2+\frac{1}{2} g t_2^2\) ….(2)

and for the third stone, \(h=\frac{1}{2} g t_3{ }^2\)…(3)

In order to eliminate u, multiplying (1) by t2 and (2) by itself and then subtracting, we obtain

⇒ \(-h\left(t_2+t_1\right)=-\frac{1}{2} g\left(t_1+t_2\right) t_1 t_2 \Rightarrow h=\frac{g}{2}\left(t_1 t_2\right)\)

But from (3) \(h=\frac{1}{2} g t_3{ }^2\)

∴ \(t_3{ }^2=t_1 t_2 \Rightarrow t_3=\sqrt{t_1 t_2}\)

Question 68. Water drops fall at regular intervals from a tap in the roof. At an instant when a drop is about to leave the tap, the separation between three successive drops is in the ratio

  1. 1:2:3
  2. 1:4:9
  3. 1:3:5
  4. 1:5:13

Answer: 3. 1:3:5

Let the time interval between two successive drops be f.

For the first drop (being detached), h1=0.

For the second drop, \(h_2=\frac{1}{2} g t^2\)

For the third drop, \(h_3=\frac{1}{2} g(2 t)^2=\frac{g}{2}\left(4 t^2\right)\)

For the fourth drop, \(h_4=\frac{1}{2} g(3 t)^2=\frac{g}{2}\left(9 t^2\right)\)

Hence, the separations between two successive drops are

⇒ \(h_2-h_1=\frac{1}{2} g t^2, h_3-h_2=\frac{1}{2} g\left(3 t^2\right) \text { and } h_4-h_3=\frac{1}{2} g\left(5 t^2\right)\)

∴ The required ratio is \(\frac{1}{2} g t^2: \frac{1}{2} g\left(3 t^2\right): \frac{1}{2} g\left(5 t^2\right)\)

= 1:3:5.

Question 69. A balloon starts rising up from the ground with an acceleration of 1.25 m s-2. After 8 s, a stone is released from the balloon. The stone will

  1. Cover a distance of 40 m
  2. Have a displacement of 50 m
  3. Reach the ground in 4 s
  4. Begin to move down after being released

Answer: 3. Reach the ground in 4 s

The balloon starts moving upwards with an acceleration of 1.25 m s-2.
Hence, its height at t = 8s will be

⇒ \(h=\frac{1}{2} a t^2=\frac{1}{2}\left(1.25 \mathrm{~m} \mathrm{~s}^{-2}\right)(8 \mathrm{~s})^2=40 \mathrm{~m}\)

and its velocity at that instant will be

⇒ \(n=a t=\left(1.25 \mathrm{~m} \mathrm{~s}^{-2}\right)(8 \mathrm{~s})=10 \mathrm{~m} \mathrm{~s}^{-1}\)

At this instant, the stone shares the motion of the balloon and falls

freely under gravity for which

⇒ \(u=10 \mathrm{~m} \mathrm{~s}^{-1}, h=-40 \mathrm{~m} \text { and } g=-10 \mathrm{~m} \mathrm{~s}^{-2}\)

∵ \(h=u t+\frac{1}{2} g t^2 \Rightarrow-40=10 t-\frac{1}{2}(10) t^2\)

⇒ \(5 t^2-10 t-40=0 \Rightarrow t^2-2 t-8=0\)

(t- 4)(t + 2) = 0

=> t = 4 s.

Thus, the stone strikes the ground in 4 s.

Question 70. A body is thrown vertically upwards- Which of the following graphs correctly represents the variation of its velocity against time?

Motion In A Staight Line multiple choice question and answers variation of its velocity against time Q 70

Answer: 4.

Motion In A Staight Line multiple choice question and answers variation of its velocity against time Q 70.

The slope of the v-t graph is the acceleration (g). At t = 0, the velocity is maximum and positive.

It goes on decreasing (remaining positive), and at the highest point (A), v 0.

Further, it becomes negative and increases with the same acceleration.

Hence, the correct graph is given in (d).

Question 71. A parachutist after bailing out falls 50 m without any friction. When the parachute opens, it decelerates at 2 m s-2. He reaches the ground at a speed of 3 m s-1. At what height has he bailed out?

  1. 91m
  2. 182 m
  3. 293 m
  4. 111 m

Answer: 3. 293 m

When the parachutist falls freely through a height of 50 m, let his
speed be u.

So, u² = 2gh

= 2(9.8 m s-2)(50 m)

= 980 (m s-1)².

When he opens the parachute, he falls through a height of h, where

32- 980 + 2(-2)h.

∴ \(h=\frac{980-9}{4} \mathrm{~m}=242.75 \mathrm{~m} \approx 243 \mathrm{~m}\)

Hence, the height at bailing out is H = 50 m + 243 m

= 293 m.

Question 72. Two identical balls are thrown upwards with the same initial velocity of 40 m s-1 in the same vertical direction at an interval of 2 s. The balls collide at a height of (taking g = 10 m s-2)

  1. 210 m
  2. 125 m
  3. 75 m
  4. 40 m

Answer: 3. 75 m

Given that u = 40 m s-1.

Let the height at which the balls collide be h.

For the first ball, \(h=u t-\frac{1}{2} g t^2\)

and for the second ball, \(h=u(t-2)-\frac{1}{2} g(t-2)^2\)

Equating the expressions for h, we obtain

⇒ \(u(t-2)-\frac{1}{2} g(t-2)^2=u t-\frac{1}{2} g t^2\)

Simplifying, we get

2u = 2g(t-1).

Substituting the values,

2(40 m s-1) = (20 ms-1)(t – Is).

∴ The time of collision is t- 5 s.

Hence, the balls collide at a height of

⇒\(h=u t-\frac{1}{2} g t^2=(40 \times 5-5 \times 25) \mathrm{m}\)

= 75m.

Question 73. A ball is projected vertically upwards from the ground. It experiences a constant air resistance of 2 m s-2 directed opposite to the direction of its motion. The ratio of the time of ascent to the time of descent equals (assuming g = 10 m s-2)

  1. 1
  2. \(\frac{2}{3}\)
  3. \(\sqrt{\frac{4}{3}}\)
  4. \(\sqrt{\frac{2}{3}}\)

Answer: 4. \(\sqrt{\frac{2}{3}}\)

During the ascent,

⇒ \(v=0=u-(g+a) t_1 \Rightarrow t_1=\frac{u}{g+a}\)… (1)

During the descent,

⇒ \(h=0\left(t_2\right)+\frac{1}{2}(g-a) t_2{ }^2=\frac{1}{2}(g-a) t_2{ }^2 .\)

For the upward motion,

⇒ \(h=u t_1-\frac{1}{2}(g+a) t_1^2\)

Substituting for t1 from (1),

⇒ \(h=\frac{u^2}{g+a}-\frac{1}{2}(g+a) \cdot \frac{u^2}{(g+a)^2}=\frac{u^2}{2(g+a)}\)

Equating the expressions for h,

⇒ \(\frac{1}{2}(g-a) t_2^2=\frac{u^2}{2(g+a)} \Rightarrow t_2=\frac{u}{\sqrt{g^2-a^2}}\)

Hence,

⇒ \(\frac{t_1}{t_2}=\frac{u}{g+a} \cdot \frac{\sqrt{g^2-a^2}}{u}=\sqrt{\frac{g-a}{g+a}}=\sqrt{\frac{10-2}{10+2}}=\sqrt{\frac{2}{3}}\)

Question 74. A ball is projected vertically upwards from the foot of a tower. It crosses the top of the tower twice after a time interval of 8s and strikes the ground after 16 seconds. The height of the tower is (taking g = 10 m s-2)

  1. 140 m
  2. 240 m
  3. 100 m
  4. 200 m

Answer: 2. 240 m

The velocity of the ball while crossing the top of the tower during its
descent is

⇒ \(v=0+g t=10 \mathrm{~m} \mathrm{~s}^{-2} \times \frac{8}{2} \mathrm{~s}=40 \mathrm{~m} \mathrm{~s}^{-1}\)

The time taken to cover the height of the tower is \(\frac{1}{2}(16 s-8 s)\) = 4s.

∴ height of the tower is

⇒ \(h=u t+\frac{1}{2} g t^2=\left(40 \mathrm{~m} \mathrm{~s}^{-1}\right)(4 \mathrm{~s})+\frac{1}{2}\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(4 \mathrm{~s})^2\)

= 160 m + 80 m

= 240 m.

Question 75. When an object is shot from the bottom of a long and smooth inclined plane kept at an angle of 60° with the horizontal, it can travel a distance xx along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel a distance of x2. The ratio of x1 to x2 will be

  1. 1:√2
  2. √2:1
  3. 1:√3
  4. 1:2√3

Answer: 3. 1:√3

In Case I, work done by gravity

⇒ \(-m g h_1=-m g x_1 \sin 60^{\circ}\)

and change in KE = \(-\frac{1}{2} m v^2\)

Hence, \(m g x_1\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{2} m v^2\)

Similarly, in Case 2

⇒ \(m g x_2 \sin 30^{\circ}=\frac{1}{2} m v^2 .\)

∴ \(m g x_1\left(\frac{\sqrt{3}}{2}\right)=m g x_2\left(\frac{1}{2}\right)\)

⇒ \(\frac{x_1}{x_2}=\frac{1}{\sqrt{3}} \Rightarrow x_1: x_2=1: \sqrt{3}\)

Motion In A Staight Line multiple choice question and answers Q 75

Question 76. The speed of a swimmer in still water is 20 m s-1. The speed of the water of a river, flowing due east, is 10 m s-1. If the swimmer is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes relative to the north is given by

  1. 30° west
  2. 60° west
  3. 45° west

Answer: 1. 30° west

Given that the velocity of the swimmer relative to the river is v st = 20 m s-1 and the velocity of the river relative to the ground is urg = 10 m s-1. These are shown vectorially in the adjoining diagram.

According to the vector diagram,

⇒ \(\vec{v}_{\mathrm{sg}}=\vec{v}_{\mathrm{sr}}+\vec{v}_{\mathrm{rg}}\)

and \(\sin \theta=\frac{v_{\mathrm{rg}}}{v_{\mathrm{sr}}}=\frac{10 \mathrm{~ms}^{-1}}{20 \mathrm{~ms}^{-1}}=\frac{1}{2} \Rightarrow \theta=30^{\circ}\)

Hence, the direction of the strokes is 30° due west.

Motion In A Staight Line multiple choice question and answers velocity of the swimmer Q 76

Question 77. A particle is moving along a circular path with a constant speed of 10 m s-2. What is the magnitude of the change in velocity of the particle when it moves through an angle of 60° around the center of the circular path?

  1. 10√2 m s-2
  2. 1.m s-1
  3. 10√3 m s-1
  4. Zero

Answer: 2. 1.m s-1

The initial velocity at A is

⇒ \(\vec{u}=\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{i}\)

and the final velocity at B is

⇒ \(\vec{v}=\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(\cos 60^{\circ}\right) \hat{i}+\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(\sin 60^{\circ}\right) \hat{j}\)

⇒ \(\left(5 \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{i}+\left(5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{j}\)

∴ the change in velocity is

⇒ \(\Delta \vec{v}=\vec{v}-\vec{u}=(5-10) \mathrm{m} \mathrm{s}^{-1} \cdot \hat{i}+5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1} \cdot \hat{j}\)

Hence, the magnitude of change in velocity is

⇒ \(|\Delta \vec{v}|=|-5 \hat{i}+5 \sqrt{3} \hat{j}| \mathrm{m} \mathrm{s}^{-1}\)

⇒ \(\sqrt{25+75} \mathrm{~m} \mathrm{~s}^{-1}=10 \mathrm{~m} \mathrm{~s}^{-1}\)

Motion In A Staight Line multiple choice question and answers initial velocity Q 77

Question 78. The position (x) of a particle as a function of time (t) is given by x(t) = at + bt² -ct³, where a, b, and c are constants. When the particle attains zero acceleration, its velocity will be

  1. \(a+\frac{b^2}{2 c}\)
  2. \(a+\frac{b^2}{c}\)
  3. \(a+\frac{b^2}{3 c}\)
  4. \(a+\frac{b^2}{4 c}\)

Answer: 3. \(a+\frac{b^2}{3 c}\)

Given that x = at + bt²- ct³.

∴ velocity = \(v=\frac{d x}{d t}=a+2 b t-3 c t^2\) …..(1)

and acceleration = \(a=\frac{d v}{d t}=2 b-6 c t\) …..(2)

For a = 0, 2b-6cf = 0

⇒ \(t=\frac{2 b}{6 c}=\frac{b}{3 c}\)

From (1), the velocity at this time t is

⇒ \(v=a+2 b\left(\frac{b}{3 c}\right)-3 c\left(\frac{b}{3 c}\right)^2=a+\frac{2 b^2}{3 c}-\frac{b^2}{3 c}=a+\frac{b^2}{3 c}\)

Question 79. A helicopter is rising vertically upwards from the ground with an acceleration g starting from rest. When it reaches a height of h, a packet is dropped at t = 0. Find the time t when the packet strikes the ground.

  1. \(\sqrt[4]{\frac{2 g}{h}}\)
  2. \(\sqrt{\frac{2 h}{8}}\)
  3. \(2 \sqrt{\frac{2 h}{g}}\)
  4. \((1+\sqrt{2}) \sqrt{\frac{2 h}{g}}\)

Answer: 4. \((1+\sqrt{2}) \sqrt{\frac{2 h}{g}}\)

Let the velocity of the helicopter at tire height h be v.

∴ \(v^2=2 g h \Rightarrow v=\sqrt{2 g h} \text { (upward). }\)

For free fall of the packet from the height h,

⇒ \(-h=u t-\frac{1}{2} g t^2=\sqrt{2 g h} t-\frac{1}{2} g t^2\)

⇒ \(\left(\frac{g}{2}\right) t^2-\sqrt{2 g h} t-h=0\)

Solving for t, we get

⇒ \(t=\frac{1}{g}[\sqrt{2 g h}+\sqrt{2 g h+2 g h}]=\frac{(2+\sqrt{2}) \sqrt{g h}}{g}\)

⇒ \((1+\sqrt{2}) \sqrt{\frac{2 h}{g}}\)

Leave a Comment