WBCHSE Class 11 Physics Notes For Relative Velocity And Relative Acceleration

Relative Velocity And Relative Acceleration

Relative Velocity: When we consider the motion of a body either on the earth’s surface or close to it, it is assumed that the earth is a stationary frame of reference. The velocity of a body with respect to a stationary observer on the earth’s surface is considered as the actual velocity.

• Often we consider the motion of a body with respect to another one.
• A passenger in a moving train observes the surrounding objects in apparent motion even if they actually are stationary. When both of two trains are moving side by side, an observer in one of them would not be able to perceive the actual motion of the other—only an apparent motion relative to himself would be observed.
• In such cases, it is necessary to consider the idea of relative motion or relative velocity between two bodies where the reference point may not be at rest.
• We already know that there is nothing like absolute rest or absolute motion. So there should be nothing like absolute velocity—all observed velocities are effectively relative.

Relative Velocity Definition: Relative velocity is defined as the apparent velocity of a body with respect to another body that may be in motion.

Calculation Of Relative Velocity:

Case 1: The observer and the object move in the same direction: Let us assume that the velocity of the observer is u and that of the object is v. If they move in the same direction, the object would seem to be slower to the observer.

The relative velocity of the object with respect to the observer is, w = actual velocity of the object – actual velocity of the observer

= V— u…..(1)

Case 2: The observer and the object move in opposite directions: Let the velocity of the observer be u and that of the object moving in the opposite direction be -v.

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Here again, the relative velocity of the object with respect to the observer is,

w = actual velocity of the object – actual velocity of the observer

= – v- u = -(v+ u)….(2)

The magnitude of the relative velocity of the object, (u + v) will always be greater than the actual velocities (u or v).

Case 3: The observer and the object move in different directions: If the object and the observer are moving in different directions that are not in the same line or plane, then again equation (1) is used to find the relative velocity. However, the subtraction has to be done using vector algebra.

In general, if \(\vec{u}\) = velocity of an object,

and \(\vec{v}\) = velocity of an observer, then, the relative velocity of the object with respect to the observer is, \(\vec{w}=\vec{v}-\vec{u}\)….(3)

We may rewrite equation (3) as \(\vec{w}=\vec{v}-\vec{u}=\vec{v}+(-\vec{u})\) = the resultant of the vectors \(\vec{v}\) and —\(\vec{u}\).

If we interchange the observer and the object keeping the velocities unchanged, then the relative velocity is given by, \(\vec{w}^{\prime}=\vec{u}-\vec{v}=-(\vec{v}-\vec{u})=-\vec{w}\)

As \(\vec{w}^{\prime}=-\vec{w}\), these two relative velocities Eire eqeal but opposite. Thus, out of two trains running on parallel tracks if the second train moves forward relative to the first one, then the first train will be observed to move backward relative to the second.

Since velocity is a vector quantity, the relative velocity of a body involves a vector subtraction and can be found out by the vector addition of \(\vec{v}\) and \(-\vec{u}\).

For example, a train moving due east with a velocity u sees another train moving due west with the same velocity. Then, relative velocity = \(\vec{u}-(-\vec{u})=2 \vec{u}\)and they will appear to approach each other at twice the speed.

If both the trains are moving in the same direction, relative velocity = \((\vec{u}-\vec{u})\) = 0. The trains will appear to be at rest with respect to each other.

Magnitude And Direction Of Relative Velocity: Let us consider that a body A (for example, an observer) is moving with velocity \(\vec{u}\) along OX and B, and another body is moving with velocity \(\vec{v}\) along OY as shown. Taking O as origin, \(\overrightarrow{O A}=\vec{u}\), \(\overrightarrow{O B}=\vec{v}\) and the angle between them is ∠AOB = α.

To compute the relative velocity of B with respect to A, OA’ is drawn so that AO = OA’. Then \(\overrightarrow{O A^{\prime}}\) is the opposite vector of \(\overrightarrow{O A}\), i.e., \(\overrightarrow{O A^{\prime}}\) = –\(\vec{u}\). Completing the parallelogram OA’CB, the diagonal \(\overrightarrow{O C}\) is drawn. OC represents the resultant of \(\vec{v}\) and –\(\vec{u}\), i.e., the relative velocity of B with respect to A.

∴ \(\overrightarrow{O A}\) = \(\vec{w}\)

And OC  = \(|\vec{w}|=\sqrt{u^2+v^2+2 u v \cos \left(180^{\circ}-\alpha\right)}\)

= \(\sqrt{u^2+v^2-2 u v \cos \alpha}\)…..(4)

If the relative velocity \(\overrightarrow{O C}\) is inclined to \(\overrightarrow{O B}\) at an angle θ then,

tanθ \(=\frac{u \sin \left(180^{\circ}-\alpha\right)}{v+u \cos \left(180^{\circ}-\alpha\right)}=\frac{u \sin \alpha}{\nu-u \cos \alpha}\)….(5)

Also, since, \(\vec{w}^{\prime}=-\vec{w}\), the relative velocity of A with respect to B is represented by the vector \(\overrightarrow{C P}\).

Relative Acceleration: By similar reasoning, relative acceleration is defined as the apparent acceleration of a body with respect to another body that may be in an accelerated motion.

Let A and B be two moving objects. The rate of change of velocity of A with respect to B is called the relative acceler¬ation of A with respect to B. Actual acceleration of A is obtained when B is at rest or in uniform motion.

If the actual accelerations of A and B are \(\vec{a}_A\) and \(\vec{a}_B\) respectively, then the relative acceleration of B with respect to A

∴ \(\vec{a}_{B A}=\vec{a}_B-\vec{a}_A\)…..(1)

and relative acceleration of A with respect to B

∴ \(\vec{a}_{A B}=\vec{a}_A-\vec{a}_B\)….(2)

Relative Acceleration Example:

1. The actual acceleration of a freely falling body is the downward acceleration due to gravity (\(\vec{g}\)). Hence, two bodies, falling freely, will have a relative acceleration \(\vec{g}\)–\(\vec{g}\) = 0. They are, therefore, stationary or moving with uniform velocity with respect to each other.
2. Let a body A fall vertically downwards with an acceleration \(\vec{g}\) due to gravity. Another body B is moving horizontally with a uniform acceleration \(\vec{a}\).

Then the relative acceleration of A with respect to B is, \(\vec{g}^{\prime}=\vec{g}-\vec{a}=\vec{g}+(-\vec{a})\) = resultant of \(\vec{g}\) and –\(\vec{a}\).

This resultant \(\vec{g}^{\prime}\) is shown.

Its magnitude is, g’ = \(\sqrt{g^2+a^2} \text { (obviously, } g^{\prime}>g \text { ) }\)

g’ is inclined with the vertical at an angle, \(\theta=\tan ^{-1} \frac{a}{g}\)

For example, when a car or a train accelerates suddenly, any object hanging from its roof is tilted away from the vertical due to the inclination of the relative acceleration \(\vec{g}^{\prime}\). For any random motion of the car or of the train, the relative acceleration also changes randomly, and angle θ does not have a steady value. That is why hanging objects are often observed to be oscillating.

Relative Acceleration Numerical Examples

Example 1. A car is moving at 80 km · h^-1 towards the north. Another car is moving at 80^2 km · h^-1 towards the northwest. Find the relative velocity of the second car with respect to the first.
Solution:

The velocity of the first car, \(\vec{u}\) = 80 km • h^-1 towards north = \(\overrightarrow{A B}\) the velocity of the 2nd car, \(\vec{v}\) = 80√2 km · h^-1 towards north-west = \(\overrightarrow{A C}\).

Hence, relative velocity of the second car with respect to the \(\vec{w}=\vec{v}-\vec{u}=\vec{v}+(-\vec{u})=\overrightarrow{A C}+\overrightarrow{B A}=\overrightarrow{B C}\) (from triangle law of vector addition)

As ∠BAC = \(45^{\circ}\), we get from \(\triangle A B C\), \(w^2=u^2+v^2-2 u v \cos 45^{\circ}\)

= \((80)^2+(80 \sqrt{2})^2-2.80 \cdot 80 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\)

∴ \(w^2=80^2\left[1+(\sqrt{2})^2-2\right]=80^2\)

∴ w = 80 km · h^-1

Since, w = u, from ΔABC,

∠ACB = ∠BAC = 45°

∴ ∠ABC = 90°

Hence, the vector \(\overrightarrow{B C}=\vec{w}\) is directed towards the west.

The relative velocity of 1st car with respect to the second will also have the same magnitude but will be directed towards the east.

Example 2. Two bodies are moving such that the velocity of one is twice that of the other and they make an angle of 60° with each other. Find the relative velocity of one with respect to the other.
Solution:

Representing the velocities in the vector diagram, we get, \(\overrightarrow{A B}=\vec{u}\) = velocity of one body, \(\overrightarrow{A B}=\vec{2 u}\) = velocity of the other.

The relative velocity of the second with respect to the first, \(\vec{w}=\overrightarrow{A C}-\overrightarrow{A B}=\overrightarrow{A C}+\overrightarrow{B A}=\overrightarrow{B C}\) [from triangle law of vector addition]

From ΔABC, \(w^2=(-u)^2+(2 u)^2+2 \cdot(-u) \cdot 2 u \cos 60^{\circ}\)

= \(5 u^2-2 u^2=3 u^2\)

∴ \(w=\sqrt{3} u\)

Also from the trigonometric rule, \(\frac{2 u}{\sin \angle A B C}=\frac{w}{\sin 60^{\circ}}=\frac{\sqrt{3} u \times 2}{\sqrt{3}}\)

or, \(\sin \angle A B C=1=\sin 90^{\circ}\)

∴ \(\angle A B C=90^{\circ} \text {. }\)

Hence \(\vec{w}\), i.e., \(\overrightarrow{B C}\) is perpendicular to \(\vec{u}\), i.e., \(\overrightarrow{A B}\).

Similarly, the relative velocity of the first with respect to the second can be found. In this case, the relative velocity will be –\(\vec{w}\) as the direction will be opposite.

Example 3. At any instant of time, two ships A and B are 70 km apart along a line AB which is directed from north to south. A starts moving towards west at 25 km · h^-1 and at the same time B starts moving towards the north at 25 km · h^-1. Find the distance of closest approach between the two ships and the time required for this.
Solution:

Solution: Let us choose the origin at point A; x-axis along east; y-axis along north.

Initial position of ship A=0 and that of ship B=-70\(\hat{j}\) km

Velocity of ship A = \(-25 \hat{i} \mathrm{~km} \cdot \mathrm{h}^{-1}\) and that of ship \(B=25 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\).

After a time of t hours, the positions of the ships are,

⇒ \(\vec{r}_A=0+(-25 \hat{i}) t=-25 t \hat{i} \mathrm{~km}\)

⇒ \(\vec{r}_B=-70 \hat{j}+(25 \hat{j}) t=(25 t-70) \hat{j} \mathrm{~km}\)

Position of ship B relative to ship A, \(\vec{r}=\vec{r}_B-\vec{r}_A=25 t \hat{i}+(25 t-70) \hat{j}\)

The distance between them is r = \(|\vec{r}|\).

So, \(r^2 =(25 t)^2+(25 t-70)^2\)

= \(625 t^2+625 t^2-3500 t+4900\)

= \(50\left(25 t^2-70 t+98\right)\)

= \(50\left[(5 t-7)^2-7^2+98\right]\)

= \(50\left[(5 t-7)^2+49\right]\)

When the distance r between the ships is minimum, \(r^2\) is also minimum.

This happens when \((5 t-7)^2\), a squared quantity, is minimum, i.e., zero.

∴ 5t – 7 = 0

or, t = \(\frac{7}{5} \mathrm{~h}=84 \mathrm{~min}=1 \mathrm{~h} 24 \mathrm{~min}\)

At this instant of time, \(r^2=50[0+49]=50 \times 49=25 \times 49 \times 2\)

∴ The distance of the closest approach between the two ships, \(r_{\min }=\sqrt{25 \times 49 \times 2}=35 \sqrt{2} \mathrm{~km}\)

Example 4. A ship is moving towards the east at 10 km · h^-1. A boat is moving north of east making an angle of 30° with the north. What should be the velocity of the boat so that the boat always appears, from the ship, to move towards the north?
Solution:

Let the velocity of the ship be \(\vec{u}=\overrightarrow{O A}\) and velocity of the boat be \(\vec{v}=\overrightarrow{O B}\)

Hence, the velocity of the boat with respect to the ship, \(\vec{w}=\vec{v}-\vec{u}=\overrightarrow{A B}\)

From the figure, \(\sin 30^{\circ}=\frac{u}{v} \text { or, } \frac{1}{2}=\frac{u}{v}\)

or, \(v =2 u=2 \times 10 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { [Given, } u=10 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { ] }\)

= \(20 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Example 5. A man is in a car moving with an acceleration of 5 m · s^-2. Find the apparent value of the acceleration due to gravity and the direction of pull of the earth with respect to him.
Solution:

Let the acceleration of the man in the car be \(\vec{a}\) and acceleration due to gravity be \(\vec{g}\). Hence acceleration due to gravity’s relation to the man is, \(\overrightarrow{g^{\prime}}=\vec{g}-\vec{a}=\vec{g}+(-\vec{a})\)

g’ = \(\sqrt{g^2+a^2}=\sqrt{9.8^2+5^2}\)

= \(\sqrt{121.04}=11 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

The angle θ, which \(\vec{g}^{\prime}\) makes with \(\vec{g}\) is given by \(\tan \theta=\frac{a}{g}=\frac{5}{9.8}=0.51=\tan 27^{\circ}\)

∴ \(\theta=27^{\circ}\)

Hence, with respect to the man the earth’s pull is acting downward at an angle 27° with the vertical in the side opposite to his direction of motion.

Example 6. A lift is moving up with a constant acceleration a. A man standing on the lift, throws a ball vertically upwards with a velocity v, which returns to the thrower after a time t. Show that v = (a + g)\(\frac{t}{2}\) where g is the acceleration due to gravity.
Solution:

The downward acceleration of the ball with respect to the lift = g-(-a) = g+ a.

The initial velocity of the ball is v upward. As the ball returns to the thrower, its relative displacement is zero.

Hence, from the equation h = ut- \(\frac{1}{2}\)gt², we get,

0 = \(v t-\frac{1}{2}(g+a) t^2 \text { or, } 0=t\left\{v-\frac{1}{2}(g+a) t\right\}\)

Since, t ≠ 0,

∴ \(v-\frac{1}{2}(g+a) t=0 \quad \text { or, } v=(a+g) \frac{t}{2}\)

Example 7. A lift is moving up with an acceleration of 2 m · s^-2. A nail gets dislodged from the roof of the lift when its speed reaches 8 m · s^-1. If the height of the lift cage is 3 m, find the time taken by the nail to touch the floor of the lift.
Solution:

Relative to the lift, the initial velocity of the nail is u = 0 and it falls through a height of h = 3 m. Inside the lift, the relative acceleration due to gravity acting on the nail downwards is

g’ = \(g-(-a)=9.8-(-2)=11.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ h = \(u t+\frac{1}{2} g^{\prime} t^2 \text { or, } 3=0 \cdot t+\frac{1}{2} \times 11.8 \times t^2\)

or, \(t^2=\frac{2 \times 3}{11.8} \text { or, } t=\sqrt{\frac{6}{11.8}}=0.713 \mathrm{~s}\)

Example 8. A simple pendulum is suspended from the roof of a car moving horizontally with an acceleration of 10 m · s^-2. What will be the angle made by the pendulum in its equilibrium position with the vertical? [g = 10 m · s^-2]
Solution:

The acceleration of the car, a = 10 m · s^-2, and acceleration due to gravity g = 10 m · s^-2. The pendulum will be inclined along the direction of the apparent pull of the earth.

Let the angle made by the pendulum with the vertical be θ.

∴ tanθ = \(\frac{a}{g}\) = \(\frac{10}{10}\) = 1 or, θ = 45

Example 9. Two parallel rail lines are directed as north-south. A train X runs towards north with a speed of 15 m · s^-1 and another train Y runs towards south with a speed of 25 m · s^-1. Find

1. The velocity of Y relative to X,
2. The velocity of ground with respect to Y,
3. The velocity of a monkey running on the roof of X against its motion with a velocity of 5 m · s^-1 relative to X, as observed by a man standing on the ground.

Solution:

Let us assume the velocity from south to north is positive.

Then, the velocity of X, v_x = +15 m · s^-1

and velocity of Y, u_y = -25 m · s^-1 m

The velocity of train Y relative to train X is, v_YX = v_Y – v_X=-25-15 = -40 m · s^-1

The relative velocity of ground with respect to train Y is, v_GY = v_G – V_Y = o – (-25) = 25 m · s^-1

The relative velocity of the monkey with respect to training X is, VMX= v_M-u_X=-5m · s^-1

(where is the velocity of the monkey with respect to the ground)

∴ v_M = v_X + v_MX = 15 + (-5) = 10 m · s^-1

Example 10. A steamer is moving towards the east with a velocity u. A second steamer is moving with a velocity 2u at angle θ north of east The motion of the second steamer relative to the first is along the northeast. Show that, cosθ-sinθ = \(\frac{1}{2}\)
Solution:

Let us choose, the x-axis along the east and the y-axis along the north. Then the velocities are of the first steamer, \(\vec{v}_1=u \hat{i}\); of the second steamer, \(\overrightarrow{v_2}=2 u \cos \theta \hat{i}+2 u \sin \theta \hat{j}\)

∴ Velocity of the second steamer relative to the first is, \(\vec{w}=\vec{v}_2-\vec{v}_1=u(2 \cos \theta-1) \hat{i}+2 u \sin \theta \hat{j}\)

As \(\vec{w}\) is along the north-east, its x- and y-components are equal.

∴ u(2cosθ-1) = 2usinθ or, 2cosθ-2sinθ = 1

or, cosθ – sinθ = \(\frac{1}{2}\)

Example 11. A stone is dropped from a tower 400 m high. Simultaneously, another stone is thrown upwards from the earth’s surface with a velocity of 100 m/s. When and where would these two stones meet? (g = 9.8 m/s²)
Solution:

Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards.

∴ The acceleration of one relative to the other = g- g = 0.

At time t = 0, the downward velocities of the two stones are, respectively, 0 and -100m/s. So, relative velocity = 0 – (-100) s 100 m/s. The stones move with this relative velocity, which is uniform in the absence of any relative acceleration.

The initial distance between the stones = 400 m.

Thus, they will meet after a time t = \(\frac{400}{100}\) = 4 s.

The height through which the first stone falls in time t = 4 s is, h = \(\frac{1}{2}\)gt² = \(\frac{1}{2}\)x 9.8 x 42 = 78.4 m

∴ The stones meet after 4 s at a height of (400 – 78.4), or 321.6 m.

Example 12. A rubber bail Is thrown downwards from the top of a lower with a velocity of 14 m/s. A second ball Is dropped from the same place 1 s later. The first ball reaches the ground in 2 seconds and rebounds with the same magnitude of velocity. How much later would the two balls collide with each other?
Solution:

Height of the tower, h = Distance travelled by the first ball in 2 s (t = 0 to t = 2s)

= 14 x 2 + \(\frac{1}{2}\) x 9.8 x 2² = 47.6 m

The second ball starts at t = 1 s. From t = Is to t = 2s, this ball comes down through a height, h’ = \(\frac{1}{2}\) x 9.8 x (2 – 1)² = 4.9 m

∴ At t = 2 s , the distance between the two balls, H = h-h’ = 47.6-4.9 = 42.7 m

For the first ball, velocity at t = 2 s is, v₁ = 14 + 9.8 x 2 = 33.6 m/s

As it rebounds with the same magnitude of velocity, its upward velocity at the same instant, i.e., at t = 2 s is v’₁= -33.6 m/s.

For the second belli, at t = 1 s, the velocity u = 0.

So, at t = 2 s, the velocity, v₂ = 0 + 9.8 x (2 – 1) = 9.8 m/s

∴ Relative velocity of the second ball with respect to the first at t = 2 s is,

V = v₂-v’₁ = 9.8-(-33.6) = 43.4 m/s

Relative acceleration =g-g = 0. So, for relative motion, the velocity V is uniform. Thus, time required to travel the height of H = 42.7 m is,

t = \(\frac{H}{V}\) = \(\frac{42.7}{43.4}\) = 0.98 s-1

Therefore, the balls will collide at T = 2 + t = 2.98 s.

Example 13. An object falling freely from a height H hits an inclined plane at a height h in its trajectory. At the instant of collision, the velocity of the object changes to become horizontal. What is the value of \(\frac{h}{H}\) for which it spends maximum time to reach the ground?

Solution:

The trajectory of the object is ABC

Let AB is a free fall through a height (H-h) in time t₁.

∴ (H- h) = \(0+\frac{1}{2} g t_1^2 \quad \text { or, } t_1=\sqrt{\frac{2(H-h)}{g}}\)

BC is a projectile motion for which the vertical component of velocity initially (at B) is zero. Also for the path BC, the vertical drop is through a height h. If the time taken is t₂

h = \(0+\frac{1}{2} g t_2^2 \quad \text { or, } t_2=\sqrt{\frac{2 h}{g}}\)

Total time, t = \(t_1+t_2=\sqrt{\frac{2(H-h)}{g}}+\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2}{g}}(\sqrt{H-h}+\sqrt{h}]\)

∴ \(t^2=\frac{2}{g}[(H-h)+h+2 \sqrt{h(H-h)}]\)

= \(\frac{2}{g}[H+\sqrt{4 h(H-h)}]=\frac{2}{g}\left[H+\sqrt{H^2-(2 h-H)^2}\right]\)

For maximum t, i.e., for maximum t², the term (2h- H)² should be minimum. This minimum value is zero, as it is a square.

∴ (2h— H)² = 0 or, 2h = H or, \(\frac{h}{H}\) = \(\frac{1}{2}\)