The position of a particle related to time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. The magnitude of velocity of the particle at $t=2 \mathrm{~s}$ will be :

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Kinematics

The position of a particle related to time is given by

x = (5 t^{2} - 4 t + 5) m

. The magnitude of velocity of the particle at

t = 2 s

will be :

(A) $10 ms^{- 1}$
(B) $14 ms^{- 1}$
(C) $16 ms^{- 1}$ ✓ Correct
(D) $06 ms^{- 1}$

Answer: (C)

Solution

The velocity is the time derivative of position,

v = \frac{d x}{d t} = 10 t - 4

. At

t = 2 s

v = 10 (2) - 4 = 16

, so the magnitude of the velocity is

16 m s^{- 1}

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option

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