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The motion of a particle along a straight line is described by equation: $x = 8 + 12t - t^3$ where x is in metre and t in second. The retardation of the particle when its velocity becomes zero, is
A
12 $ms^{-2}$
B
24 $ms^{-2}$
C
zero
D
6 $ms^{-2}$
Detailed Solution
$v = \frac{dx}{dt} = 12 - 3t^2$
v = 0: $12 - 3t^2 = 0 \Rightarrow t = 2$ s
$a = \frac{d^2x}{dt^2} = -6t$
At t = 2 s: $a = -12$ $ms^{-2}$, i.e. a retardation of 12 $ms^{-2}$
v = 0: $12 - 3t^2 = 0 \Rightarrow t = 2$ s
$a = \frac{d^2x}{dt^2} = -6t$
At t = 2 s: $a = -12$ $ms^{-2}$, i.e. a retardation of 12 $ms^{-2}$