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Equations of Motion is the basic concept that is used to describe the state of motion of any object. These equations of motion can be applied to an object in 1D, 2D, or 3D planes. There are three equations of motion that form the basis of the Physics’ mechanics and are also used to find the Displacement, acceleration, velocity, and initial velocity.
Key Takeaways: Distance, displacement, speed, velocity, time, acceleration, initial velocity, final velocity, average velocity, average speed, gravity, uniform acceleration, uniform motion
Important Definitions and Formulas
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Let’s revise the basics of the terms that will be used in deriving the equations.
| Term | Definition | Formula/ Denoted as |
| Motion | Motion can be described as the change in the position of object with respect to time | Velocity, speed, and acceleration are quantities used to describe an object in motion |
| Distance | The sum of the distance that an object covers at the given time / The actual measure of the total change in the position of an object at a particular time is called distance. It is a scalar quantity, thus only has magnitude and lacks direction. | ‘d’ – distance |
| Displacement | The shortest measure of the net change in the position of the object in a specific time interval is known as the displacement of the object It is a vector quantity, that is distance and direction change of the object are taken into consideration. | ‘s’ – displacement |
| Speed | The ratio of the distance travelled by an object (regardless of its direction) to the time required to travel that distance is known as speed. Speed can say ‘how fast or slow’ an object is in motion. Since the quantities used to derive speed are scalar, speed is also a scalar quantity. | Speed = dt Where d is distance and t is time taken SI Unit – m/s |
| Velocity | The rate of change of displacement of an object with respect to time, or the rate of change of position, is called velocity Since displacement is used to measure velocity, it is a vector quantity. | Velocity ‘v’ = st Where s is displacement and t is time taken SI Unit – m/s |
| Acceleration | The change in velocity of an object per unit time, or the rate of change of velocity function with respect to time, is defined as acceleration. | Acceleration ‘a’ = vt = (v-u)/ t Where v is velocity, u is initial velocity, and t is time taken SI Unit – m/s2 |
Distance
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Equations of Motion by Graphical Method
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The equation of motion helps us to understand the behaviour and nature of a physical system in uniform acceleration (constant acceleration with motion occurring in a straight line). These equations are used to relate and find out quantities like displacement, distance, initial velocity, final velocity, time, and acceleration of the object. The three equations of motion are:
| Equation for velocity-time relation | v = u + at |
| Equation for position (displacement)-time relation | s = ut + 12 (at2) |
| Equation for position (displacement)-velocity relation | v2 = u2 – 2as |
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Derivations of Equations
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Derivations of equation of motion are as follows.
Velocity-Time Relation
Consider the velocity-time graph of an object that moves under uniform acceleration as shown in Figure 1, here initial velocity is not equal to 0 (u ≠ 0). From this graph, you can see that the initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t. The velocity changes at a uniform rate a.
Figure 1
In Figure 1, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC.
Let us draw AD parallel to OC. From the graph, we observe that
BC = BD + DC = BD + OA
Substituting BC = v and OA = u, we get
v = BD + u
or BD = v – u … (1)
From the velocity-time graph (Figure – 1), the acceleration of the object is given by
a = Change in velocity time taken
= BDAD = BDOC
Substituting OC = t, we get
a = BDt
or BD = at … (2)
Using Eqs. (1) and (2) we get
v = u + at
Read More: Radius of Gyration
Position-Time Relation
Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Figure - 1, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
Thus, the distance s travelled by the object is given by
s = area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABD
= OA × OC + 12 (AD × BD) … (3)
Substituting OA = u, OC = AD = t and BD = at, we get
s = u × t + 12 (t × at)
or
s = u t + 12 at2 … (4)
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Position-Velocity Relation
From the velocity-time graph shown in Figure - 1, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is,
s = area of the trapezium OABC
= OA + BC ×OC 2
Substituting OA = u, BC = v and OC = t, we get
s = u+v t 2 … (5)
From the velocity-time relation (4), we get
t = v–u a … (6)
Using Eqs. (5) and (6) we have
s = (v+u) (v-u) 2a
or 2as = v2 – u2
v2 = u2 – 2as
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Things to Remember
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- The slope of the velocity vs time graph gives the value of acceleration, similarly, the slope of the distance-time graph gives the value of speed, and the slope of the displacement-time graph gives the value of the velocity of the object in motion.
- We can easily represent uniform or non-uniform motion through graphs and thus identify them.
- If the motion of an object is traced as a straight line parallel to the time axis in a distance-time graph, the object is stationary.
- In a distance-time graph, an object in uniform motion shows a straight line with constant slope while an object in non-uniform motion will have a curve line with increasing-decreasing slopes.
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Sample Questions
Ques. The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h–1 and m s–1. [2 marks]
Ans. Distance covered by the car,
s = 2400 km – 2000 km = 400 km
Time elapsed, t = 8 h
Average speed of the car is,
vav = st= 400 km 8 h = 50 km h–1
= 50 km/h × 1000 m 1km × 1h 3600s
= 13.9 m s–1
The average speed of the car is 50 km h–1 or 13.9 m s–1.
Ques. Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases [4 marks]
Ans. In the first case:
initial velocity, u = 0
final velocity, v = 6 m s–1
time, t = 30 s
a = v -u /t
Substituting the given values of u,v and t in the above equation, we get
a = 6-030
a = 0.2 m s–2
In the second case:
initial velocity, u = 6 m s–1;
final velocity, v = 4 m s–1;
time, t = 5 s.
Then, a = 4-65
a = –0.4 m s–2.
The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second case, it is –0.4 m s–2.
Ques. Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.[2 marks]
Ans. Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
Average speed = Total distance covered Total time taken
= 180m 1min = 180m 1min × 1min 60s = 3 m s–1
Average velocity = Displacement Total time taken
= 0m 60s
= 0 m s–1
The average speed of Usha is 3 m s–1 and her average velocity is 0 m s–1.
Ques. A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the acceleration is uniform, find [3 marks]
(a) the accelerationbr
(b) the distance travelled by the train for attaining this velocity.
Ans. We have been given u = 0 ; v = 72 km h–1 = 20 m s–1 and t = 5 minutes = 300 s.
- We know that
a = Ê‹-ut
a = 20-0300
a = 115 ms-2
- We have 2as = v2 – u2= v2– 0
Thus,
s = v22a
s = 20×202×(1/15)
= 3000 m
= 3 km
The acceleration of the train is 1 15 m s– 2 and the distance travelled is 3 km.
Ques. A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate
(a) the acceleration
(b) the distance covered by the car at that time.[3 marks]
Ans. We are given that
u = 18 km h–1 = 5 m s–1
v = 36 km h–1 = 10 m s–1
and t = 5 s.
- We have a = Ê‹-ut
a = 10-55
a = 1 m s–1
- We have s = u t + 12 at2
= 5 m s–1 × 5 s + 12 × 1 m s–2 × (5 s) 2
= 25 m + 12.5 m
= 37.5 m
The acceleration of the car is 1 m s–2 and the distance covered is 37.5 m
Ques. The brakes applied to a car produce an acceleration of 6 m s–2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.[2 marks]
Ans. We have been given
a = – 6 m s–2
t = 2 s
and v = 0 m s–1.
We know that v = u + at
0 = u + (– 6 m s–2) × 2 s
or u = 12 m s–1.
We get s = u t + 12 at2
= (12 m s–1) × (2 s) + 1 2 (–6 m s–2) (2 s)2
= 24 m – 12 m
= 12 m
Thus, the car will move 12 m before it stops after the application of brakes.
Ques. A bus starts from rest and moves with constant acceleration 8ms−2. At the same time, a car travelling with a constant velocity 16 m/s overtakes and passes the bus. After how much time and at what distance, the bus overtake the car?[3 marks]
Ans. Let the position of the bus be PB and the position of the car be PC.
From s = ut + 12 at², we have,
The initial velocity of the bus, u = 0, hence we have PB = 12 (8)t²
And PC = Velocity × Time
= 16 × t.
For the bus to overtake the car, we must have: PB = PC
Hence, 12 (8)t² = 16 × t.
Therefore, t = 4s.
Using the value of t = 4s in PB = 12 (8)t ², we have the position of the bus at the time of overtaking is = 64m.
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