Introduction: Motion in a Straight Line - Science & Technology for UPSC CSE PDF Download

Did you know that everything in the universe is always moving? Even if you are sleeping on your bed, you still belong to the Earth which is continuously moving about its axis and around the Sun. This means that you are also moving. It is interesting to note that nothing in this universe is at complete rest. Therefore it is important to understand the basic ideas of motion and the quantities used to describe it.

In this document we study motion in a straight line and the basic kinematic quantities that describe that motion: distance, displacement, speed, velocity and acceleration. We also study frames of reference, uniformly accelerated motion, motion under gravity, instantaneous quantities and relative motion in one dimension.

What is Frame of Reference?

A frame of reference is an abstract coordinate system together with a set of physical reference points that fix that coordinate system and standardise measurements within it. Motion (for example, velocity) is meaningful only after we specify the frame of reference from which the motion is described.

Illustration:

In the figure above, an observer A attached to B may say that B is at rest. An observer C, however, may see B moving with velocity V in the positive x-direction. Hence, before specifying the velocity we must specify the frame of reference.

Types of Frames of Reference

• Inertial frame of reference:
  A frame of reference in which Newton’s first law of motion holds true. A body remains at rest or in uniform straight-line motion unless acted upon by a net external force. Frames moving with constant velocity relative to each other are inertial frames.
• Non-inertial frame of reference:
  A frame of reference in which Newton’s first law does not hold. Such a frame is accelerating with respect to an inertial frame, and fictitious (pseudo) forces like centrifugal or Coriolis forces appear.

Types of Frame of References

Point Mass Object

In many problems an extended object can be treated as a point mass when the dimensions of the body are negligible compared with the distances involved in the motion. This simplifies analysis while keeping results accurate for kinematics and dynamics.

A satellite orbiting around Earth is a Point Mass Object

• An example is a satellite orbiting Earth: although the satellite has size and shape, for orbital calculations its size is negligible compared to orbital radius, so it is treated as a point mass.
• Treating an object as a point mass simplifies calculations of trajectory, speed and gravitational interactions.

Motion and Rest

• Motion: An object is said to be in motion if its position changes with time relative to a chosen frame of reference. Motion is described by quantities such as speed, velocity and acceleration.

Motion

• Rest: An object is at rest if its position does not change with time relative to the chosen frame of reference. Rest and motion are relative: an object at rest in one frame may be in motion in another.
• Relativity of rest and motion: There is no absolute state of rest. For example, a person sitting in a house is at rest with respect to Earth but in motion with respect to the Moon or the Sun.

Distance and Displacement

Important

• If a particle is moving in a straight line without a change in direction, the magnitude of displacement is equal to the distance traveled, otherwise, displacement is always less than distance. Thus, the numerical ratio of displacement to distance is always less than or equal to one. 
• The value of displacement is always zero in case the initial point and final point of a motion are the same.

Example 1: An object moves along a grid through points A, B, C, D, E and F as shown. The side of each square tile measures 0.5 km. (Figure below)

Sol: The distance covered by the moving object is the length of the total path covered
AB + BC + CD + DE + EF
3 + 1 + 1.5 + 0.5 + 0.5 = 6.5 km
The distance covered by the moving object is 6.5 km.

Displacement (magnitude):

Applying the Pythagorean formula, we get, 
(AF)2 = (AH)2 + (HF)2   
= (4 x 0.5)2 + (3 x 0.5)2 = (2)2 + (1.5)2 = 4 + 2.25 = 6.25 km 
AF = 2.5 km 

Example 2: A particle moves from A to B along a circular arc of radius R subtending an angle θ at the centre. Find the ratio of distance to magnitude of displacement.

Ans:

The distance traveled by the particle is the length of arc AB. 
Arc length = radius x theta = Rθ 
Displacement = AB 
If we drop a perpendicular bisector from the center of the circle on AB we will get as shown in the figure, 

From the figure above we can conclude that, displacement = AB = 2Rsin θ/2

Example 3: A wheel of radius 1 m rolls forward half a revolution on horizontal ground. What is the magnitude of displacement of the point on the rim that was initially in contact with the ground?

Sol: 

Speed and Velocity

Speed

Speed is the rate of change of position irrespective of direction. It is a scalar quantity.

• S.I. unit: m s^-1.
• Dimensional formula: [M⁰ L¹ T^-1].

(a) Uniform speed

If an object covers equal distances in equal intervals of time it has uniform speed (uniform motion).

Example 4: A car travels 300 km in 4 hours. Calculate its speed.

Sol:

Distance = 300 km.

Time = 4 h.

Speed = Distance / Time.

Speed = 300 km / 4 h = 75 km h^-1.

Thus the uniform speed is 75 km h^-1.

This means that the car is traveling at a constant rate of 75 kilometers for every hour of travel time.

(b) Average speed

The ratio of the total distance traveled by the object to the total time taken is called the average speed of the object.

• If a particle travels distances s1, s2, s3, … with speeds v1, v2, v3, …, then, 
  
• If particle travels equal distances (s1 = s2 = s) with velocities v1 and v2, then, Average 
  
• If a particle travels with speeds v1, v2, v3, …, during time intervals t1, t2, t3,…, then, Average speed = (v1t1 + v2t2 + v3t3 +…) / (t1 + t2 + t3 +…)
• If particle travels with speeds v1, and v2 for equal time intervals, i.e.,   t1 = t2 = t3, then, Average speed = v₁ + v₂/2
• When a body travels an equal distance with speeds v1 and v2, the average speed (v) is the harmonic mean of two speeds, 2/v = 1/v₁ + 1/v₂
  

Example 5: A car travels 60 km at 40 km h^-1 and then another 60 km at 60 km h^-1. What is the average speed?

Sol:

Distance for first segment = 60 km.

Speed for first segment = 40 km h^-1.

Time for first segment = 60 / 40 = 1.5 h.

Distance for second segment = 60 km.

Speed for second segment = 60 km h^-1.

Time for second segment = 60 / 60 = 1.0 h.

Total distance = 60 + 60 = 120 km.

Total time = 1.5 + 1.0 = 2.5 h.

Average speed = Total distance / Total time = 120 / 2.5 = 48 km h^-1.

Example 6: A particle moves in a straight line covering half the distance with speed v₀. The remaining part is covered in the remaining time with speeds v₁ and v₂ each for half of that remaining time. Find the average speed over the whole motion.

Sol:

Given that
A particle covered the journey from A to D = 2s
It covered A to B i.e. = s
With speed =v₀
Time taken 

It covered B to D i.e. = 2s
In time = t
Such that
We get
BD = BC + CD

Example 7: A man walks 2.5 km to a market at 5 km h^-1. Finding it closed, he returns and walks for the remaining 10 minutes at 7.5 km h^-1. Total travel time is 40 minutes. Find the average speed for the whole journey.

Sol:Distance from home to market = 2.5 km

Distance from market back to home = 2.5 km

∴ Total distance travelled = 2.5 + 2.5 = 5 km

Total time of travel = 40 min = 40/60 h = 2/3 h

Average speed = Total distance / Total time

Average speed = 5 / (2/3)
= 5 × 3/2
= 15/2
= 7.5 km/h

∴ Average speed of the man for the whole journey = 7.5 km/h

(c) Non-uniform or variable speed

When an object covers unequal distances in equal time intervals its speed is non-uniform (variable). Such motion requires average or instantaneous descriptions.

(d) Instantaneous Speed

The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous speed. It is the rate of change of distance with respect to time.
Instantaneous speed is always greater than or equal to zero and is a scalar quantity. For uniform motion, instantaneous speed is constant. To understand it in simple words we can say that instantaneous speed at any given time is the magnitude of instantaneous velocity at that time. It is a limit of the average speed as the time interval become very small.
A moving object does not have the same speed during its travel. Sometimes it speeds up and sometimes slows down. At a given instant time what we read from the speedometer is instantaneous speed. When a cop pulls you over for speeding, he clocked your car’s instantaneous speed or speed at a specific point in time as your car sped down the road.

Example 8: A car starts from rest, increases its speed uniformly to 20 m s^-1 over 100 m, then decelerates uniformly to rest over 50 m. Find the average speed for the entire motion.

Sol: Given:

Initial speed, u₁ = 0 m s^-1

Final speed, v₁ = 20 m s^-1

Distance, s₁ = 100 m

Initial speed for second part, u₂ = 20 m s^-1

Final speed, v₂ = 0 m s^-1

Distance, s₂ = 50 m

Total distance = 100 + 50 = 150 m

Time for first part:

Using   v = u + at

20 = 0 + a t₁   ⇒   a = 20 / t₁

Using   s = ut + ½ a t²

100 = 0 + ½ (20 / t₁) t₁²

100 = 10 t₁   ⇒   t₁ = 10 s

Time for second part:

Using   v = u + at

0 = 20 + a t₂   ⇒   a = −20 / t₂

Using   s = ut + ½ a t²

50 = 20 t₂ − ½ (20 / t₂) t₂²

50 = 20 t₂ − 10 t₂ = 10 t₂

∴ t₂ = 5 s

Total time = t₁ + t₂ = 10 + 5 = 15 s

Average speed = Total distance / Total time

Average speed = 150 / 15 = 10 m s^-1

∴ Average speed for the entire motion = 10 m s^-1

Velocity

Velocity is the rate of change of displacement in a particular direction. It is a vector quantity (magnitude and direction).

• Unit: m s^-1.
• Dimensional formula: [M⁰ L¹ T^-1].
• Velocity may be positive, zero or negative (sign indicates direction in chosen frame).

Example 9: A car covers 150 km in 2 h. Find its velocity (assuming straight-line motion in a fixed direction).

Sol: Displacement = 150 km.

Time = 2 h.

Velocity = Displacement / Time = 150 / 2 = 75 km h^-1.

Example 10: A car travels from A to B in 3 h and returns to A in 5 h. A and B are 150 miles apart along a straight highway. What is the average velocity for the whole trip?

Sol: The velocity of the car is zero in this case as displacement is zero (initial and final positions are the same). 

(a) Uniform Velocity
If an object undergoes equal displacements in equal intervals of time, then it is said to be moving with a uniform velocity. 

Important
This is only possible in the case where a body is traveling in a straight line without changing its direction.

(b) Non-Uniform or Variable Velocity
If an object undergoes unequal displacements in equal intervals of time, then it is said to be moving with a non-uniform or variable velocity.

Important: This is possible only if the body changes the magnitude of speed or its direction w.r.t. Time.

(c) Average Velocity 
The ratio of the total displacement to the total time taken is called average velocity.

(d) Instantaneous Velocity
The rate of change of displacement of an object in a particular direction is its velocity. Its S.I unit is meter per second.

The direction of instantaneous velocity at any time gives the direction of motion of a particle at that point in time. The magnitude of instantaneous velocity equals the instantaneous speed. This happens because, for an infinitesimally small time interval, the motion of a particle can be approximated to be uniform.

Example 11: Suppose a hiker starts at point A and walks 10 kilometers to point B in 2 hours. After a short break at point B, the hiker continues walking for another 6 kilometers to reach point C in 1.5 hours. To find the average velocity for the entire journey, we can use the formula:
Sol: Average velocity = total displacement / total time taken.
The total displacement is the sum of the individual displacements, which is 10 kilometers + 6 kilometers = 16 kilometers. 
The total time taken is the sum of the individual times, which is 2 hours + 1.5 hours = 3.5 hours.
Plugging these values into the formula:
average velocity = 16 km / 3.5 h ≈ 4.57 km/h.
Therefore, the average velocity for the entire journey is approximately 4.57 kilometers per hour.

Example 12: Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Sol: (a) From A to B
Total distance covered from A to B = 300 m
Total displacement = 300m
Total time taken = 2 × 60 s + 50 s = 170 s

Therefore, average speed from A to B = Total Distance/Total Time = 300/170 = 1.76ms⁻¹
Average velocity from A to B = Displacement AB/Time = 1.76ms⁻¹ 
(b) From A to C
Total distance covered from A to C = AB + BC = 300 + 100 m = 400 m
Total time taken from A to C = = 170 s + 60 s = 230 s
Therefore, average speed from A to C = Total Distance/Total Time = 400/230⁻¹ =1.74ms⁻¹ 
Displacement from A to C = AC= AB - BC = 300 - 100 = 200 m
Time taken for displacement from A to C = 230 s
Therefore, average velocity from A to C = Displacement /Time = 200/230ms⁻¹ = 0.87ms⁻¹ 

Example 13: An insect crawls a distance of 4m along the north in 10 s and then a distance of 3m along the East in 5 s. What is the average velocity of the insect?
Sol: 

Important

Often we use the terms speed and velocity interchangeably without realizing it is wrong to do so. There is a difference between speed and velocity which everyone must know. Below is the difference between these two terms.

Acceleration

Acceleration is the time rate of change of velocity. It is a vector quantity.

• Unit: m s^-2.
• Dimensional formula: [M⁰ L¹ T^-2].
• Acceleration may be positive (speeding up), zero (uniform velocity), or negative (retardation or deceleration).

Average Acceleration

It is the change in velocity divided by an elapsed time. For instance, if the velocity of a marble increases from 0 to 60 cm/s in 3 seconds, its average acceleration would be 20 cm/s². This means that the marble’s velocity will increase by 20 cm/s every second.

It is the rate of change of velocity with respect to displacement

Acceleration is a = dv/dt

∴ a = dv/(dx/v)

a = v(dv/dx)

If a particle experiences accelerations a₁ for time t₁ and a₂ for time t₂ then the average acceleration is a_av = (a₁t₁ + a₂t₂) / (t₁ + t₂).

Acceleration of the Velocity-Time graph

In the given graph, a = (40-20)/(4-2) = 10 m/s². For a particle it is equal to the slope of a velocity-time graph.

Example 14: A car starts from rest and reaches 30 m s^-1 in 6 s under uniform acceleration. Find average acceleration.

Sol:

Initial velocity u = 0 m s^-1.

Final velocity v = 30 m s^-1.

Time t = 6 s.

Average acceleration = (v - u) / t = (30 - 0) / 6 = 5 m s^-2.

Motion in Different Acceleration for Different Time Intervals

Let’s understand this through an example. Suppose, a particle started its motion from rest with an acceleration of 1m/s² for 2s and then continued it for next 1s changing to 2m/s². The distance travelled during this will be:

After 2s the velocity is, v = u+at = 2 m/s

Now, if this is the initial velocity for the second half of the motion, s2=ut+(1/2)at² =3 m

Distance traveled in first half is: s₁ = 0+(1/2)at² = 2 m

Hence total distance traveled = s₁+s₂ = 5 m

Uniformly Accelerated Motion

Uniformly accelerated motion refers to the motion of an object that is experiencing a constant acceleration. In this type of motion, the object's velocity changes at a constant rate over time.

Equations of Motion

If a body starts with velocity (u) and after time t its velocity changes to v, if the uniform acceleration is a and the distance traveled in time t in s, then the following relations are obtained, which are called equations of uniformly accelerated motion.
(i) v = u + at

• v represents the final velocity of an object.
• u represents the initial velocity of an object.
• a represents the constant acceleration experienced by the object.
• t represents the time taken for the object to undergo the acceleration.

(ii) s = ut + 1/2 at²

• s represents the displacement or distance traveled by an object.
• u represents the initial velocity of the object.
• a represents the constant acceleration experienced by the object.
• t represents the time taken by the object.

(iii) v² = u² + 2as

• v represents the final velocity of an object.
• u represents the initial velocity of the object.
• a represents the constant acceleration experienced by the object.
• s represents the displacement or distance traveled by the object.

(iv) Distance traveled in nth second Sn = u + a/2 (2n – 1)

• Sn represents the distance traveled during the nth second.
• u represents the initial velocity of the object.
• a represents the constant acceleration experienced by the object.
• n represents the time in seconds, specifically the nth second of motion.

If a body moves with uniform acceleration and velocity changes from u to v in a time interval, then the velocity at the midpoint of its path.

Here are some numerical examples for the above equations: 
(i) v = u + at:
Example 15: Let's say a car starts from rest (initial velocity u = 0) and accelerates at a rate of 2 m/s2 for 5 seconds. We can calculate the final velocity (v) using the equation:
v = u + at
v = 0 + (2 m/s2)(5 s) = 10 m/s
Therefore, the final velocity of the car after 5 seconds of acceleration is 10 m/s.
(ii) s = ut + 1/2at²:
Example 16: Consider a ball that is thrown upwards with an initial velocity of 20 m/s. If we want to find the displacement (s) of the ball after 3 seconds, we can use the equation:
s = ut + 1/2at² 
s = (20 m/s)(3 s) + 1/2(-9.8 m/s²)(3 s)² ≈ 15 m
Hence, the ball has a displacement of approximately 15 meters after 3 seconds.

(iii) v² = u² + 2as:
Example 17: Suppose a body starts from rest (initial velocity u = 0) and experiences a constant acceleration of 4 m/s². If we want to find the final velocity (v) after covering a distance of 100 meters, we can use the equation:
v2 = u2 + 2as
v2 = 02 + 2(4 m/s2)(100 m) = 800 m2/s2
Taking the square root of both sides, we get:
v ≈ 28.28 m/s
Therefore, the final velocity of the body after covering a distance of 100 meters is approximately 28.28 m/s.

(iv) Distance traveled in nth second Sn = u + a/2(2n – 1):
Example 18: Let's say an object is moving with an initial velocity of 10 m/s and an acceleration of 2 m/s2. We want to find the distance traveled during the 4th second. We can use the formula:
Sn = u + a / 2(2n – 1)
S₄ = 10 m/s + (2 m/s2) / 2(2(4) – 1)
S₄ = 10 m/s + 1 (7)
S₄ = 17 m
Hence, the distance traveled during the 4th second is 17 meters.

Motion Under Gravity

The force of attraction of earth on bodies is called the force of gravity. Acceleration produced in the body by the force of gravity is called acceleration due to gravity. It is represented by the symbol g. In the absence of air resistance, it is found that all bodies (irrespective of size, weight, or composition) fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small altitude (h<<R) is called free fall. (1) If a body drops from some height, taking an initial position as the origin and direction of motion (i.e., downward direction) as a positive, here we have:

If the initial velocity is zero, then u = 0.
Graph of distance velocity and acceleration with respect to time:

(3) If a body is projected vertically upward, taking an initial position as origin and direction of motion (i.e., vertically up) as positive then, 
a=-g [As acceleration is downwards while motion upwards]
So, if the body is projected with velocity u and after time t it reaches up to height s then,
 
Graph of distance, velocity, and acceleration with respect to time (for maximum height):

It is clear that both quantities do not depend upon the mass of the body or we can say that in the absence of air resistance, all bodies fall on the surface of the earth at the same rate. Here are some numerical examples for the above equations:

Example 19: Freely falling object (u = 0) under gravity:
(i) v = u + gt:
Consider an object falling freely under gravity with an acceleration of g = 9.8 m/s². If we want to find the final velocity (v) after 3 seconds of free fall, we can use the equation:
v = u + gt
v = 0 + (9.8 m/s2)(3 s)
v = 29.4 m/s
(ii) h = ut + 1/2gt2:
To determine the height (h) traveled by the object during the 3 seconds of free fall, we can use the equation:
h = ut + 1/2gt2
h = 0 + 1/2(9.8 m/s2)(3 s)2
h = 44.1 m
(iii) v2 = u2 + 2gh:
Using the equation to find the square of the final velocity (v) of the freely falling object:
v2 = u2 + 2gh
v2 = 02 + 2(9.8 m/s2)(44.1 m)
v2 = 862.68 m2/s2
v ≈ 29.38 m/s

Example 20: Object thrown upward: If an object is thrown upward, the acceleration due to gravity (g) is replaced by (-g) in the equations of motion. Let's consider the same initial conditions and time values as in Example 1:
(i) v = u + (-g)t:
Using the equation for the final velocity (v) of the object thrown upward:
v = u + (-g)t
v = 0 + (-9.8 m/s2)(3 s)
v = -29.4 m/s
(ii) h = ut + 1/2(-g)t2:
To calculate the height (h) reached by the object thrown upward during the 3 seconds, we use the equation:
h = ut + 1/2(-g)t2
h = 0 + 1/2(-9.8 m/s2)(3 s)2
h = -44.1 m
(Note: The negative sign indicates that the object is below the starting point.)
(iii) v2 = u2 + 2(-g)h:
Applying the equation for the square of the final velocity (v) of the object thrown upward:
v2 = u2 + 2(-g)h
v2 = 02 + 2(-9.8 m/s2)(-44.1 m)
v2 = 862.68 m2/s2
v ≈ 29.38 m/s
These examples illustrate how the equations of motion can be used to analyze the motion of freely falling objects and objects thrown upward under the influence of gravity.

Non- Uniformly Accelerated Motion

Non-uniformly accelerated motion refers to the motion of an object that is changing its velocity at a non-constant rate. In other words, the object is accelerating or decelerating, but the acceleration itself is not constant over time. This type of motion can be more complex to analyze than uniformly accelerated motion, where the acceleration remains constant. To understand this kind of motion, one must know about Instantaneous velocity.

When an object is traveling with variable velocity, its speed at a given instant of time is called its instantaneous velocity.

Example 21: Let's say a car is traveling along a straight road, and its position is given by the equation: x(t) = 3t2 - 2t + 1, where x is the position of the car at time t. Find the instantaneous speed of the car at a specific time t.
Sol: To find the instantaneous speed of the car at a specific time t, we need to calculate the derivative of the position function with respect to time. The derivative will give us the rate of change of position, which is the velocity.
Differentiating x(t) with respect to t, we get:

Now, to find the instantaneous speed at a particular time, substitute the value of t into the velocity function.
Let's say we want to find the instantaneous speed at t = 2 seconds:
v(2) = 6(2) - 2 = 12 - 2 = 10 m/s.
Therefore, the instantaneous speed of the car at t = 2 seconds is 10 m/s.

Graphs Related To Kinematics

(i) Uniform Motion - This type of motion is defined as the motion of an object in which the object travels in a straight line and its velocity remains constant along that line as it covers equal distances in equal intervals of time, irrespective of the duration of the time.  Example: If the speed of a car is 10 m/s, it means that the car covers 10 meters in one second. The speed is constant in every second.



(ii) Non Uniform Motion - This type of motion is defined as the motion of an object in which the object travels with varied speed and it does not cover same distance in equal time intervals, irrespective of the time interval duration. Example: If a car covers 10 meters in first two seconds, and 15 meters in next two seconds.

Relative Motion in 1-D

The velocity of an object only has meaning when it is expressed relative to a set of axes or a frame of reference.

We usually consider the motion of objects relative to the Earth, but since the Earth is orbiting the Sun and the Sun is moving through the Galaxy, this is a purely arbitrary concept. When dealing with more than one body in motion, the problem becomes complex — consider the difficulties in sending the space probe Giotto to rendezvous with Halley's Comet many millions of miles away!

We should always specify the frame of reference relative to which an object is moving.

The relative velocity of one object with respect to another object is the time rate of change of the relative position of one object with respect to another object.
The relative velocity of object A with respect to object B
VAB = VA – VB

• When two objects are moving in the same direction, then
  VAB = VA - VB
  or
  VAB = VA - VB
  
• When two objects are moving in the opposite direction, then
  VAB = VA + VB
  or
  VAB = VA + VB
  

Consider the case where both bodies are moving relative to a third body. A simple example is when the motions of both bodies are in the same straight line — for instance, two cars travelling along a motorway. If both cars are travelling in the same direction, one at 25 ms^-1 and the other at 35 ms^-1, then their relative velocity is 10 ms^-1 (found by vector subtraction).

If they are moving in opposite directions, however, the relative velocity of one car with respect to the other is 60 ms^-1 (See Figure 1).

What we are effectively doing is considering one car to be at rest and finding the velocity of the other car in that frame of reference. To do this, we add the negative of the velocity of one car to both cars' velocities. This effectively brings one car to rest, and we then consider the velocity of the other car relative to it.

• When two objects are moving at an angle, then
  
  and tan β = vB sin θ / vA – vB cos θ

The situation becomes a little more complex when the motion of the two objects is not along the same straight line. Consider the case shown in Figure 2.

Here the two cars are still moving at the same speeds but this time at right angles to each other as they approach a junction.

The relative velocity is found using a vector triangle.

If we add the negative of the velocity of the red car (−25 ms^-1) to both the velocity of the red car and the green car, we can imagine the red car to be at rest and then find the velocity of the green car relative to it (See Figure 3).

This is most easily done by drawing the vector diagram. The resultant shown by the vector R is the velocity of the green car relative to the red car. In our example, the velocity of the green car relative to the red car is 43 ms^-1 at a direction of 325^o.
Think about this result and see if it matches your common-sense view of how you would see the green car moving if you were in the red car. It would seem to be moving up the page and towards the left, which fits the result exactly!

Relative motion at angles other than 90^o
We will now consider the case where the cars are not moving at right angles to each other.

In fact, we will not consider cars but two aeroplanes shown in Figure 4 (a).
To find the velocity of plane A relative to plane B, we must 'make plane B stationary'. We do this by adding the negative of plane B's velocity to both planes. If we were on plane B, it would be like making that plane stop, and we would then see plane A moving relative to us. This is shown in Figure 4(b).

Example 22: Let's consider two cars, Car A and Car B, traveling in the same direction along a straight road. Car A is moving at a speed of 60 km/h, while Car B is moving at a speed of 80 km/h.
Sol: To determine the relative velocity of Car B with respect to Car A, we subtract the velocity of Car A from the velocity of Car B.
Relative velocity of Car B with respect to Car A = Velocity of Car B - Velocity of Car A
Relative velocity = 80 km/h - 60 km/h = 20 km/h.
Therefore, the relative velocity of Car B with respect to Car A is 20 km/h. This means that Car B is moving 20 km/h faster than Car A in the same direction.

Summary