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JEE · NEET Previous Year Question

$The velocity of a particle is given by v = 2 t^{2} - 8 t + 15 ms^{- 1} . Find its instantaneous acceleration at t = 5 s .$

Topic: JEE PYQ

Answer: D. Full step-by-step solution below.

A $18 ms^{- 2}$
B $20 ms^{- 2}$
C $5 ms^{- 2}$
D $12 ms^{- 2}$

Correct answer: D

Solution

(d) Velocity, v = 2 t^{2} - 8 t + 15 a = \frac{dv}{dt} = \frac{d}{dt} (2 t^{2} - 8 t + 15) a = 4 t - 8 a at (t = 5) = (4 \times 5) - 8 = 20 - 8 = 12 m / s^{- 2}

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