JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Kinematics

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A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10 m in t s, the distance travelled by the toy in the next t s will be :

(A)10 m
(B)20 m
(C)30 m✓ Correct
(D)40 m

Answer: (C)

Solution

The toy starts from rest (u = 0) with constant acceleration a.

For the first t seconds: Using

s = u t + \frac{1}{2} a t^{2}

, we get

10 = \frac{1}{2} a t^{2}

.

For the total time 2t seconds: The total distance is

s_{t o t a l} = \frac{1}{2} a (2 t)^{2} = 2 a t^{2} = 4 \times \frac{1}{2} a t^{2} = 4 \times 10 = 40

m.

Distance in the next t seconds =

40 - 10 = 30

m.

Answer: (C) 30 m

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option

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