Chapter 3.8 - Relative velocity in Rectilinear motion

In the previous section we saw the three equations of motion. In this section we will see relative velocity.
1. Consider ourselves being seated in a train moving with uniform velocity. 
2. Let another train move parallel to our train. Also let it move in the same direction as our train. 
3. If that train overtakes us, it means that, that train is moving faster than ours. 
• But we feel that, that other train is moving slowly
4. If that other train is moving with the same uniform velocity as ours, we will feel that, that other train is not moving at all. 
• But for a person standing on the ground, both the trains will appear to be moving with the same speed.
■ Such observations can be explained by using the concept of relative velocity.

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• Relative velocity can be easily understood with the help of graphs.
• We have seen the details about position-time graph in a previous section. 
• Here, we are going to see such a graph. But this time, the distances covered by two objects will be shown in a single graph. Consider fig.3.52 below:

Fig.3.52

1. Two vehicles are travelling with the same uniform velocity of 7 ms^-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
    ♦ The coordinates of P are (4,68)
    ♦ The coordinates of Q are (4,118)
3. So we can write:
• At the instant when the stop watch reading is '4 s', vehicle 1 is at a distance of 68 m from the origin O  
• At the instant when the stop watch reading is '4 s', vehicle 2 is at a distance of 118 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (118-68) = 50 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph

    ♦ The coordinates of R are (10,110)

    ♦ The coordinates of S are (10,160)
6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 110 m from the origin O  
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 160 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (160-110) = 50 m
(same as before)
8. So we can write:
At the instant when t = 4, the distance between the two vehicles is 50 m
At the instant when t = 10, the distance between the two vehicles is 50 m
9. We can check at any instant. We will find that, the distance between the two vehicles is always 50 m. This is shown in the table below:

10. So, for a person seated in vehicle 1, the other vehicle 2 will always appear to be at a distance of 50 m
■ So, for that person, the vehicle 2 will appear to be stationary
11. The reverse is also true:
For a person seated in vehicle 2, the other vehicle 1 will always appear to be at a distance of 50 m
■ So, for that person, the vehicle 1 will appear to be stationary

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Now we will consider another case. See fig.3.53 below:

Fig.3.53

1. Two vehicles are travelling with uniform velocities
• The uniform velocity of vehicle 1 is 5 ms^-1
• The uniform velocity of vehicle 2 is 9 ms^-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
    ♦ The coordinates of P are (4,50)
    ♦ The coordinates of Q are (4,106)
3. So we can write:
• At the instant when the stop watch reading is '4', vehicle 1 is at a distance of 50 m from the origin O  
• At the instant when the stop watch reading is '4', vehicle 2 is at a distance of 106 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (106-50) = 56 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph

    ♦ The coordinates of R are (10,80)

    ♦ The coordinates of S are (10,160)
6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 80 m from the origin O  
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 160 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (160-80) = 80 m
8. So we can write:
• At the instant when t = 4, the distance between the two vehicles is 56 m
• At the instant when t = 10, the distance between the two vehicles is 80 m
9. So as time increases, the distance between the two vehicles go on increasing.
• We can check at any instant. The table below gives more evidence.

• When t = 0, distance between the two vehicles is 40 m
• When t = 1, distance between the two vehicles is 44 m. This is an increase of 4 ms^-1
• When t = 2, distance between the two vehicles is 48 m. This is a further increase of 4 ms^-1
• When t = 3, distance between the two vehicles is 52 m. This is a further increase of 4 ms^-1
• When t = 4, distance between the two vehicles is 56 m. This is a further increase of 4 ms^-1
So on . . .
10. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be moving further and further away
• The distance increasing by 4 m every second
• Thus, for the person seated in vehicle 1, the other vehicle will appear to be moving at a velocity of 4 ms^-1.
11. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving away in the opposite direction. 
• That is., vehicle 1 will appear to be moving with a velocity of -4 ms^-1  
12. As time passes, the distance between the two vehicles become so large that, they will no longer be able to see each other
•  It is same as, one vehicle is stationary and the other is moving away at an uniform velocity of 4 ms^-
1.

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Is there an easy method to obtain this 'velocity of 4 ms^-1' ?
Let us try:
• Both the vehicles are travelling in the same direction. 
    ♦ Which is: towards the positive side of the x axis (This is clear from the rising graphs)
• Vehicle 1 is travelling at 5 ms^-1 
• Vehicle 2 is travelling at 9 ms^-1
■ We find that '4' is the difference between the two individual velocities 
Let us write a general form:
1. We know the equation for each of the yellow lines. It is: x = x₀ + vt
    ♦ x is the displacement at any time t
    ♦ x₀ is the y intercept, which is the displacement when t = 0
    ♦ v is the uniform velocity with which the object is travelling
• This equation gives the displacement at any time 't'
2. So, for an object A, moving with an uniform velocity of v_A, the displacement from the origin at any time 't' is given by: x_A = x₀ + v_At
• Similarly, for an object B, moving with an uniform velocity of v_B, the displacement from the origin at any time 't' is given by: x_B = x₀ + v_Bt
3. Then at any particular instant when the stop watch shows 't', the displacement of object B with respect to object A 
= (x_B - x_A) = [(x₀ + v_Bt) - (x₀ + v_At)] 
⟹ (x_B - x_A) = (v_B - v_A)t
Consider the quantity (v_B - v_A) in the above equation
• This is in the familiar form: Distance = velocity × time
4. So at any instant, when viewed from object A, the other object B will appear to be moving with a velocity of (v_B - v_A) ms^-1.   
■ We say that: Relative velocity of B with respect to A is given by:
Eq. 3.6: v_BA = (v_B - v_A) ms^-1
• Note the order in which the subscripts are written.
• In our present case, if we consider the vehicles 1 and 2 as objects A and B respectively, we get:
Relative velocity of vehicle 2 with respect to vehicle 1 = v_BA = (9 - 5) = 4 ms^-1.
• That is., when viewed from vehicle 1, the other vehicle 2 will appear to be moving with a velocity of 4 ms^-1.
5. In a similar way,
• Relative velocity of A with respect to B is given by:
v_AB = (v_A - v_B) ms^-1.
• In our present case:
Relative velocity of vehicle 2 with respect to vehicle 1 = v_AB = (5 - 9) = -4 ms^-1.
• That is., when viewed from vehicle 2, the other vehicle 1 will appear to be moving with a velocity of -4 ms^-1. The -ve sign indicates the opposite direction. 

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Now we will consider one more case. See fig.3.54 below:

Fig.3.54

• Note that, in the above graph, in the x axis, 1 unit = 2 s
• In the previous graphs, 1 unit was 1 s
1. Two vehicles are travelling with uniform velocities in opposite directions
• The uniform velocity of vehicle 1 is 6 ms^-1
• The uniform velocity of vehicle 2 is -3 ms^-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
    ♦ The coordinates of P are (4,44)
    ♦ The coordinates of Q are (4,148)
3. So we can write:
• At the instant when the stop watch reading is '4', vehicle 1 is at a distance of 44 m from the origin O  
• At the instant when the stop watch reading is '4', vehicle 2 is at a distance of 148 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (148-44) = 144 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph

    ♦ The coordinates of R are (10,80)

    ♦ The coordinates of S are (10,130)
6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 80 m from the origin O  
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 130 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (130-80) = 50 m
8. So we can write:
• At the instant when t = 4, the distance between the two vehicles is 144 m
• At the instant when t = 10, the distance between the two vehicles is 50 m
9. So as time increases, the distance between the two vehicles go on decreasing.
• We can check at any instant. The table below gives more evidence.
• Note that, in the time column, the interval between individual entries is 2 s
• In the previous tables, it was 1 s.
This change is made to accommodate more entries

■ We notice some interesting points:
• When t = 15.55 s, the distance between the two vehicles is zero
• That means, as time increases, the distance between the two vehicles goes on decreasing to such a level that, at a particular point, the distance between them becomes zero
• It is clear that, at the instant when the distance is zero, both the vehicles are at the same point. It is the meeting point between the two vehicles. It is indicated by the point M in the graph
• Also it is highlighted in yellow colour in the table.
• But remember that, the vehicles are travelling in opposite directions. So, as time increases further, the vehicles will move away from each other.
Let us consider the travel upto the point M
• When t = 0, distance between the two vehicles is 140 m
• When t = 2, distance between the two vehicles is 122 m. This is a decrease of 18 ms^-1
• When t = 4, distance between the two vehicles is 104 m. This is a further decrease of 18 ms^-1
• When t = 6, distance between the two vehicles is 86 m. This is a further decrease of 18 ms^-1
• When t = 8, distance between the two vehicles is 68 m. This is a further decrease of 18 ms^-1
So on . . .
10. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be coming closer and closer
• The distance decreasing by 18 m every 2 seconds
• '18 m every 2 seconds' is '9 m every one second'
• Thus, for the person seated in vehicle 1, the other vehicle will appear to be coming closer and closer with a velocity of 9 ms^-1.
11. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving in the opposite direction. 
• That is., vehicle 1 will appear to be coming closer and closer with a velocity of -9 ms^-1  
12. As time passes, the distance between the two vehicles falls to zero. That is the meeting point
13. Let us consider the travel after the point M
• When t = 16, distance between the two vehicles is 4 m
• When t = 18, distance between the two vehicles is 22 m. This is an increase of 18 ms^-1
• When t = 20, distance between the two vehicles is 40 m. This is a further increase of 18 ms^-1
• When t = 22, distance between the two vehicles is 58 m. This is a further increase of 18 ms^-1
• When t = 24 distance between the two vehicles is 76 m. This is a further increase of 18 ms^-1
So on . . .
14. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be moving away
• The distance increasing by 18 m every 2 seconds
• '18 m every 2 seconds' is '9 m every one second'
• Thus, after point M, for the person seated in vehicle 1, the other vehicle will appear to be moving away with a velocity of 9 ms^-1.
15. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving in the opposite direction. 
• That is., vehicle 1 will appear to be moving away with a velocity of -9 ms^-1  
16. As time passes, the distance between the two vehicles become so large that, they will not be able to see each other.

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Is there an easy method to obtain this 'velocity of 9 ms^-1' ?
Let us try:
• Vehicle 1 is travelling towards the positive side of the x axis
(This is clear from the rising graph)
• Vehicle 2 is travelling towards the negative side of the x axis
(This is clear from the falling graph)
That is why we wrote earlier:
• The uniform velocity of vehicle 1 is 6 ms^-1
• The uniform velocity of vehicle 2 is -3 ms^-1
■ We find that '9' is the difference between the two individual velocities (∵ 6 - (-3) = 9)
Let us write a general form. The steps are similar to those in the previous case.
• At any particular instant when the stop watch shows 't', the displacement of object B with respect to object A 
= (x_B - x_A) = [(x₀ + v_Bt) - (x₀ + v_At)] 
⟹ (x_B - x_A) = (v_B - v_A)t
So Eq.3.6 is valid here also

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In the next section, we will see a special case of relative velocity.

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Chapter 3.7 - Equations of Motion

In the previous section we saw that:
■ The area enclosed between a velocity-time graph and the x axis will give the displacement of the object
• Based on that information, we will now derive the following:
Kinematic equations for uniformly accelerated motion
• That is., when objects move with uniform acceleration, certain equations can be applied to find various quantities like displacement, velocity, time duration etc.,
• We will derive those equations now
1. In a previous section, we derived Eq.3.1: v = v₀ + at
• We know that it is the equation of the velocity-time graph for an object moving with uniform acceleration.
• Such a graph is shown below:

Fig.3.50

2. From the graph, the following points are clear:
• When the stop watch was turned on, the velocity of the object was 'v₀'.
• When the stop watch showed time as 't' seconds, the velocity of the object was 'v'
3. Knowing these facts, we can calculate the 'area of the shaded portion'.
• For easy calculation, we split the area into a rectangle and a triangle
• Area of the rectangle = v₀ × t
• Area of the triangle = (¹⁄₂ × base × altitude) = (¹⁄₂ × t × (v-v₀))
• So total area = [v₀ × t + (¹⁄₂ × t × (v-v₀))]
4. Now consider Eq.3.1 that we derived earlier: v = v₀ + at.
• Rearranging this, we get: (v-v₀) = at
• Substituting this in (3), we get:
Total area = [v₀ × t + (¹⁄₂ × t × (at))]
⟹ Total area = v₀t + ¹⁄₂ × at².
5. But this total area is the displacement 'x' in the duration between the following instances:
• Instance when the stop watch showed t₁ = 0 s    
• Instance when the stop watch showed t₂ = t s
Obviously, the duration = (t₂ - t₁) = (t - 0) = t s
• So we can write:
Eq.3.2: x = v₀t + ¹⁄₂ × at²
6. Now consider the result in (3) again: 
Total area = [v₀ × t + (¹⁄₂ × t × (v-v₀))] 
• But 'total area' is 'x'. So we can write: 
x = [v₀ × t + (¹⁄₂ × t × (v-v₀))]
• This can be rearranged as shown below:

■ So we can write:
Displacement = average velocity × time 
7. Now we can write a comparison between two items:
(i) Motion with uniform velocity
• There is only one value for velocity
• Multiplying this velocity by time will give the displacement x  
(ii) Motion with uniform acceleration
• Velocity changes continuously with time
• Multiplying 'average velocity' by time will give the displacement x
8. Now consider again Eq.3.1 that we derived earlier: v = v₀ + at.
• Rearranging it, we get: t = ^(v-v₀)⁄_a
• Substituting this in (6), we get: x = [^(v+v₀)⁄₂ × ^(v-v₀)⁄_a] = [^(v2-v₀2)⁄_2a]
Rearranging, we get:
Eq.3.3: v² = v₀² + 2ax

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So we have derived 3 equations. Let us write them in order:
Eq.3.1: v = v₀ + at
Eq.3.2: x = v₀t + ¹⁄₂ × at²
Eq.3.3: v² = v₀² + 2ax
• These 3 equations connect the five quantities: v₀, v, a, t and x
• They are the kinematic equations of rectilinear motion with uniform acceleration
• Note that, they are applicable only when the object moves with uniform acceleration

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We will see some solved examples:
Click here for Solved examples 3.2, 3.3, 3.4 and 3.5 

Motion of object under free fall

1. Consider an object released from the top of a tall building or a cliff
• Since it is 'simply released', it's initial velocity v₀ will be zero
2. The object will fall towards the ground. It will be under a constant acceleration
• This acceleration is the 'acceleration due to gravity'
• It is denoted by the symbol 'g' and has a value of 9.8 ms^-2
3. We assume that, the object is under the same acceleration of 9.8 ms^-2 during it's entire fall
■ But this is true only if the 'height of fall' is small compared to the 'radius of the earth'
Why is it so?
• Ans: The acceleration due to gravity is caused due to the pull by the earth towards it's centre.
• This pull will be different if the object is at a large distance from the earth
• So g cannot be given the same value of 9.8 ms^-2, for the entire fall if the 'height of fall' is very large  
• For the problems in this chapter, we can take g = 9.8 ms^-2

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We will see some solved examples:
Click here for Solved examples 3.6, 3.7, 3.8 and 3.9 .

Stopping distance of vehicles

• When brakes are applied to a moving vehicle, it will not come to rest immediately. 
• It will come to rest only after moving a certain distance. So we can write the following points
1. Consider the two instances:
(i) Instant at which brakes are applied
(ii) Instant at which the vehicle comes to rest
2. There will be a time duration between these two instances
■ The vehicle will move a certain distance in this duration
3. It is important to find this distance for safety purposes. Let us try:
• Let the vehicle be moving with a velocity of v₀
• Let the 'application of brakes' produce a negative acceleration of '-a'
• Here acceleration is taken as negative because it is in a direction opposite to the direction of motion
4. Let the vehicle move a distance of 'd_s' in the duration mentioned in (5) above
Then 'd_s' is called the stopping distance
• Applying Eq.3.3 we get: v² = v₀² + 2ax.
⟹ 0² = v₀² + 2(-a)d_s =
⟹ v₀² = 2ad_s
⟹ Eq.3.4: d_s = [^v₀2⁄_2a]
5. The significance of d_s is shown in the fig.3.51(a) below:

Fig.3.51

(i) The car moves with a initial velocity of v₀ 
(ii) At the instant when it reaches P, brakes are applied
(iii) The car will come to rest only after travelling a distance of d_s from P
(iv) If we measure ds from P, we will be able to mark the point Q at which the car comes to rest
(v) In fig.a, we see that, Q falls before the obstacle present on the road  
(vi) So the car will not crash into the obstacle
• However, it may be noted that, if the obstacle moves towards the car, then the above d_s will not be sufficient
• Vehicle manufactures and High way designers do extensive research and calculations taking into account all the factors, to ensure safety
• It is the responsibility of the drivers to keep the speed below accepted limits  
6. Now let us see what happens if we increase the speed:
• If the car travels with double the velocity as in the above case, we can write:
Initial velocity = 2v₀
• Substituting in Eq.3.4, we get: d_s = [^(2v₀)2⁄_2a] = [^4v₀2⁄_2a]
• This new value of d_s is 4 times the previous value. That is., the stopping distance increases by 4 times
• This is shown in fig.3.51(b)
(i) The car moves with a initial velocity of 2v₀ 
(ii) At the instant when it reaches P, brakes are applied
(iii) The car will come to rest only after travelling a distance of 4d_s from P
(iv) If we measure 4d_s from P, we will be able to mark the point Q at which the car comes to rest
(v) In fig.b, we see that, Q falls after the obstacle present on the road  
(vi) So the car will  crash into the obstacle
■ We can write: Smaller the velocity, smaller will be the 'required d_s'

Reaction time

• If some incident happens suddenly with out any pre notice, we will need some time to respond to that incident.
• For example, while driving a car, if an obstacle appears suddenly on the road, some time will pass before the driver slams the brakes. This 'elapsed time' is called reaction time.
• It is the time taken to observe, think and act
• Reaction time depends on the complexity of the situation and also on the individual
• We can effectively use the 'acceleration due to gravity' to measure the reaction time of any individual. Let us see how it is done:
1. Drop a small object with out any pre notice
2. The individual whose reaction time is to be determined, must catch the object before it hits the ground
3. Then we want the distance between the following two points:
(i) Point of release
(ii) Point of catch
• This distance is 'h', the height of fall
• If we use a meter scale as the falling object, we can easily measure 'h'
4. Then we apply Eq.3.2: x = v₀t + ¹⁄₂ × at².
• Substituting the values, we get:
h = 0 + ¹⁄₂ × gt².
⟹ t2 = ^2h⁄_g
⟹ Eq.3.5: t = √[^2h⁄_g].
• This 't' is the time required by the object to fall from the 'point of release' to the 'point of catch'
• So this 't' is the reaction time.

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In the next section, we will see relative velocity.

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