First slide

Instantaneous acceleration

Question

The motion of a particle along a straight line is described by equation $x = 8 + 12 t - t^{3}$ where, x is in meter and t in second. The retardation of the particle when its velocity becomes zero, is

Moderate

A

$24 {ms}^{- 2}$
B

zero
C

$6 {ms}^{- 2}$
D

$12 {ms}^{- 2}$

Solution

Given, $x = 8 + 12 t - t^{3}$
We know that, $v = \frac{d x}{d t}$ and $a = \frac{d v}{d t}$
So, $v = 12 - 3 t^{2}$
and t = 2s,
v = 0
and $a = - 12 {ms}^{- 2}$
So, retardation of the particle = $12 {ms}^{- 2}$ .

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