JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Kinematics

The displacement and the increase in the velocity of a moving particle in the time interval of t to

(t + 1)

s are 125 m and 50 m/s, respectively. The distance travelled by the particle in

(t + 2)^{th}

s is ______ m.

Answer: 175

Solution

With constant acceleration, the velocity increase in 1 s gives

a = 50 m/ s^{2}

.

Displacement in that 1 s:

125 = u (1) + \frac{1}{2} (50) (1)^{2} \Rightarrow u = 100 m/s

(velocity at time

t

).

Distance in the

n

th second after

t

, with

n = 2

S = u + \frac{a}{2} (2 n - 1) = 100 + 25 (3) = 175 m

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Integer answer

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