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2.Motion in Straight Line
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A particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 -3t^2 + 2)\,m$ The displacement when the acceleration becomes zero is........$m$
A
$0$
B
$2$
C
$3$
D
$-2$
Solution
$s=t^{3}-3 t^{2}+2$
$\therefore $ $\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^{2}-6 \mathrm{t}$
and acceleration $\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{dt}^{2}}=6 \mathrm{t}-6$
when acceleration is zero, $\mathrm{t}=1 \mathrm{sec}.$
Hence, displacement of the particle at $t=1$ $sec$
(i.e., when acceleration becomes zero) is,
$ s =t^{3}-3 t^{2}+2 $
$=1-3+2=0 \mathrm{\,m} .$
Standard 11
Physics
