## Chapter 1 Motion Long Answer Type Question And Answers

**Question 1. The given figure shows the velocity-time graph of a bus travelling along a straight road for 24 minutes. Describe its motion. Find**

**its retardation****total distance travelled by the bus**

**Answer.**

**Given:**

The given figure shows the velocity-time graph of a bus travelling along a straight road for 24 minutes.

(1) Acceleration a = \(\frac{v-u}{t}\)

= \(\frac{(0-25) \mathrm{kmhr}^{-1}}{4 \text { minutes }}\)

= \(-\frac{\left(25 \times \frac{5}{18}\right)}{4 \times 60 \mathrm{~s}} \mathrm{~ms}^{-1}\)

= \(-\frac{125}{18 \times 4 \times 60} \mathrm{~ms}^{-2}\)

= −0.0289 ms^{-2}

∴ Retardation of the bus = 0.0289 ms^{-2}

(2) Total distance travelled = Area of OSQP + Area of ∆QSR

**Read and Learn More NEET Foundation Long Answer Questions**

= \((N P \times N S)+\frac{1}{2} \times S R \times Q S\)

= \(25 \mathrm{kmhr}^{-1} \times 18 \mathrm{~min}+\frac{1}{2} \times 25 \times 6\)

= \(\begin{aligned}

& \left(25 \times \frac{5}{18}\right) \mathrm{ms}^{-1} \times(18 \times 60) \mathrm{s} \\

& +\frac{1}{2} \times\left(25 \times \frac{5}{18}\right) \mathrm{ms}^{-1} \times 6 \times 60 \mathrm{~s}

\end{aligned}\)

= 25 × 5 × 60 + 25 × 5 × 10

= 7500 + 1250

= 8750 m

**Question 2. Barrel of a rifle is 60 cm long. A bullet initially at test, when fired from the rifle, leaves the barrel with a velocity of 300 ms−1. Find**

**Acceleration of the bullet****Time taken by the bullet to emerge from the rifle**

**Answer. **

**Given:**

Barrel of a rifle is 60 cm long. A bullet initially at test, when fired from the rifle, leaves the barrel with a velocity of 300 ms−1.

Given u = 0

v = 300 ms^{-1}

S = 60 cm = 0.60 m

a = ?

t = ?

(1) ∵ v^{2} = u^{2} +2aS

300^{2} = 0 + 2 × a × 0.6

∴ 1.2 a = 300 × 300

a = \(\frac{300 \times 300}{1.2}\)

= 300 × 250

∴ a = 7.5 × 104 ms^{-2}

(2) ∵ v = u + at

300 = 0 + 7.5 × 10^{4}t

∴ 7.5 × 10^{4} t = 300

∴ t = \(\frac{300}{7.5 \times 10^4}\)

∴ t = 4 × 10^{-3} s

**Question 3. When a truck was travelling on a horizontal road with 72 km/hr, its brakes failed. It is found that the truck comes to a complete halt after travelling 100 m on the road. How much retardation is produced in the truck by the road friction?**

**Answer. **

**Given:**

When a truck was travelling on a horizontal road with 72 km/hr, its brakes failed. It is found that the truck comes to a complete halt after travelling 100 m on the road.

Given u = 72 km/hr^{-1}

= 72 × \(\frac{5}{18} \mathrm{~ms}^{-1}\)

= 20 ms^{-1}

v = 0

S = 100 m

a = ?

∵ v^{2} = u^{2} +2aS

0 = (20)^{2 }+ 2 × a × 100

∴ 200 a = −400

∴ a = −2 ms^{-2}

∴ Retardation of the truck = 2 ms^{-2}

**Question 4. In a cathode ray tube, an electron travels from rest from cathode to anode, a distance of 50 cm. When it reaches the anode, it has a velocity of 3 × 107 ms ^{-1}. Find**

- Acceleration of the electron
- Time it takes to reach the anode.

**Answer. **

**Given:**

In a cathode ray tube, an electron travels from rest from cathode to anode, a distance of 50 cm.

Given u = 0

v = 3 × 107 ms^{-1}

S = 50 cm = 0.50 m

a = ?

t = ?

(1) ∵ v^{2} = u^{2} + 2aS

(3 × 107)^{2} = 0 + 2 × a × 0.5

9 × 10^{14} = a

∴ a = 0.9 × 1015 ms^{-2}

(2) ∵ v = u + at

v = at (∵ u = 0)

∴ t = \(\frac{v}{a}=\frac{3 \times 10^7}{9 \times 10^{14}}=0.33 \times 10^{-7} \mathrm{~s}\)

**Question 5. A train 150 m long is running with a constant speed of 54 km/hr. How long will it take to cross**

- a signal?
- a platform 300 m long?

**Answer. **

**Given:**

A train 150 m long is running with a constant speed of 54 km/hr.

Speed of the train = 54 km/hr^{-1}

= 54 × \(\frac{5}{18}\)

= 15 ms^{-1}

(1) To cross a signal means to travel a distance equal to its length.

∵ Time = \(\frac{\text { Distance }}{\text { Speed }}=\frac{l}{v}=\frac{150 \mathrm{~m}}{15 \mathrm{~ms}^{-1}}=10 \mathrm{~s}\)

(2) To cross a platform the train has to cover a

distance = L + l = 300 m + 150 m

= 450 m

∵ Time = \(\frac{S}{v}=\frac{450 \mathrm{~m}}{15 \mathrm{~ms}^{-1}}=30 \mathrm{~s}\)

**Question 6. An express train (E) and a goods train (G) are travelling with constant speeds of 45 ms ^{-1} and 15 ms^{-1} respectively. By mistake, they were on same track and approaching each other. When they were 500 m apart, drivers of both the trains realise the mistake and immediately applied brakes to produce retardation of 3 ms^{-2} and 2 ms^{-2 }respectively. Will they be able to avert accident?**

**Answer.**

**Given:**

An express train (E) and a goods train (G) are travelling with constant speeds of 45 ms^{-1} and 15 ms^{-1} respectively. By mistake, they were on same track and approaching each other. When they were 500 m apart, drivers of both the trains realise the mistake and immediately applied brakes to produce retardation of 3 ms^{-2} and 2 ms^{-2 }respectively.

Distance S_{1 }travelled by express train before coming to rest can be found out using:

v^{2} = u^{2} +2as

0 = (45)^{2} + 2(−3)S_{1}

6 S_{1} = 2025

∴ \(S_1=\frac{2025}{6}=337.5 \mathrm{~m} .\)

Similarly distance travelled by goods train before coming to rest is given by

v^{2} = u^{2} +2as

⇒ 0 = 15^{2} + 2(−2)S_{2}

⇒ 4 S_{2 }= 225

∴ S_{2} = 56.25 m

Total distance travelled by them before coming to rest S = S_{1} + S_{2} = 337.5 m + 56.25 m

= 393.75 m

Since they were initially 500 m apart, there are still (500 − 393.25) m = 106.25 m apart.

So, there is no accident.

**Question 7. Elaborate distance–time graph with a diagram.**

**Answer. **

**Distance–time graph with a diagram:**

In order to calculate the velocity, consider points A and B on the diagonal line. S_{1} and S_{2} are the points on y-axis where A and B points touch horizontally. T_{1} and T_{2} are the points on x-axis where A and B touch vertically.

Distance = BC

= S_{2} – S_{1}

Time = AC

= T_{2} – T_{1}

AB = \(\frac{B C}{A C}\)

v = \(\frac{\left(S_2-S_1\right)}{\left(T_2-T_1\right)}\)

Where v = velocity

(S_{2} – S_{1}) = Interval of distance

(T_{2} – T_{1}) = Interval of time

Therefore, Velocity = \(\frac{\text { Distance }}{\text { Time }}\)

**Question 8. Deduce the equation for**

**Velocity – Time relation****Position – Time relation****Position – Velocity relation**

**Answer. **

For the object moving with a uniform acceleration, the following three equations give the relationship between initial velocity (v), acceleration (a), time of journey (t) and distance travelled (S).

(1) **Equation for Velocity-Time relation:**

Consider the above figure: The initial velocity of an object is u which gradually increases to v in time t. The graph shows uniform velocity.

Initial velocity (at t = 0) = OA = u

Final velocity (at t = t) = OC = v

Acceleration a = Slope of the line AB

a = EB/AE

= AC/OD

= (OC – OA)/OD

= (v – u)/t

at = v – u

v = u + at

(2) **Equation for Position–Time Relation:**

Consider the above figure:

Distance S travelled in time t = Area of trapezium OABD

= Area of rectangle OAED + Area of triangle ABE

= \(O A \times O D+\frac{1}{2} \times B E \times A E\)

= \(u \times t+\frac{1}{2} \times(v-u) \times t\)

As we know

v – u = at

Therefore,

S = \(u t+\frac{1}{2} a t^2\)

(3)** Equation for Position-Velocity Relation:**

Consider the above figure:

Distance S travelled in time t = Area of trapezium OABD

= \(\frac{1}{2}(O A+D B) \times O D\)

= \(\frac{1}{2}(u+v) \times t\)

As we know

t = (v – u)/a

Therefore,

S = \(\frac{1}{2}(u+v)\left(\frac{v-u}{a}\right)\)

= \(\frac{1}{2}\left(\frac{v^2-u^2}{a}\right)\)

2aS = v^{2} – u^{2}

v^{2} = u^{2}+ 2aS

**Question 9. The distance between the house and the school of a girl is 3.6 km. If she takes 6 minutes to reach then her school by car, calculate her speed in m/s. Also express her speed in km/h.
**

**Answer.**

**Given:**

The distance between the house and the school of a girl is 3.6 km.

Distance, S = 3.6 km = 3.6 × 103m

Time, t = 6 min = 6 × 60 = 360 s = 6/60 h

Speed, v = ?

Using formula

v = S/t

v = 3.6/6/60

v = 3.6 × 60/6

v = 36 km/h

Also,

v = 3.6 × 10^{3}/360

v = 10 m/s