Equation Of Motion, Types Of Equation, Important Topics For JEE Main 2024

First Equation of Motion : The basic quantities of motion such as displacement, velocity, acceleration, and time of an object can be described using the equations of motion. Here, let us discuss the derivations of the equations of motion and find the displacement in the nth second using the equations of motion.

First Equation of Motion

In the figure, the velocity of the object at times t ₁ and t ₂ are u and v respectively. The time interval for the motion is, t = t ₂ – t ₁ . Here u is the initial velocity and v is the final velocity. We know that the slope of a velocity-time graph will give us the object's acceleration.

Hence, we can find the slope here.

Slope = a = (v-u)/(t ₂ – t ₁ ).

i.e.,

Second Equation of Motion

Second Equation of Motion : From the above figure, we can find the displacement of the object by finding the area under the graph. Here,

Displacement,  s = Area of the trapezium ABDE

s = 1 2 × sum of parallel sides x perpendicular height

s = 1 2 × (v + u) t

Third Equation of Motion

Third Equation of Motion : From the above figure, we can find the displacement of the object by finding the area under the graph. Here,

s = × (v + u)

s =

Equations of motion using Calculus

Equation of a freely falling object : For a freely falling object, the value of acceleration will be acceleration due to gravity (g) and the initial velocity will zero and displacement will be equal to the height covered.

v = gt

h = gt ²

Equation of Motion Examples

Q.1. A particle having initial velocity 10 m/s moves with a constant acceleration 5 ms ^–2 , for a time 15 second along a straight line, what is the displacement of the particle in the last 2 second?

Ans.

Q.2: A particle moving with a uniform acceleration travels 24 m and 64 m in the first two consecutive intervals of 4 sec each. Its initial velocity is:

1. 1m/sec 2. 10m/sec 3. 5m/sec 4. 2m/sec

Ans: A → B

24 = 4 u + 8 a

6 = u + 2 a

B → C ….. (i)

24 + 64 = 4(8) + a × 8 ²

88 = 8 u + 32 a

11 = u + 4 a

Eqn (i) and (ii)

u = 1 m / s

Q.3. The initial velocity of a particle is 10 m/s retardation is 2 m/s². The distance covered in the fifth.

second of the motion will be:

Ans. u = 10 m/s;

a = –2 m/s ²

Displacement