The displacement of a particle is represented by the following equation s = 3t³ + 7t² + 5t + 8 where s is in metres t in seconds. The acceleration of the particle at t = 1 s is

Q. The velocity

v

of a particle moving along a straight line and its distance

s

from a fixed point on the line are related by

v^{2} = a^{2} + s^{2}

, then its acceleration equals

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A particle moves along a straight line according to the law s=16−2t+3t3, where s metres is the distance of the particle from a fixed point at the end of t second. The acceleration of the particle at the end of 2 s is

The correct option is D 36m/s2Given, s=16−2t+3t3⇒dsdt=−2+9t2⇒d2sdt2=18tNow the acceleration of the particle at the end of t=2s isf=d2sdt2=18×2=36m/s2

A particle moves along a straight line according to the law $s = 16 - 2 t + 3 t^{3}$ , where $s$ metres is the distance of the particle from a fixed point at the end of $t$ second. The acceleration of the particle at the end of $2$ s is

The correct option is D $36 m / s^{2}$
Given, $s = 16 - 2 t + 3 t^{3}$
$\Rightarrow \frac{d s}{d t} = - 2 + 9 t^{2}$
$\Rightarrow \frac{d^{2} s}{d t^{2}} = 18 t$
Now the acceleration of the particle at the end of $t = 2$ s is
$f = \frac{d^{2} s}{d t^{2}} = 18 \times 2 = 36 m / s^{2}$