Motion in a Straight Line - Result Question 29
30. A particle moves along a straight line such that its displacement at any time $t$ is given by $s=(t^{3}-6 t^{2}+3 t+4)$ metres
The velocity when the acceleration is zero is
(a) $3 ms^{-1}$
(b) $-12 ms^{-1}$
(c) $42 ms^{-2}$
(d) $-9 ms^{-1}$
[1994]
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Answer:
Correct Answer: 30. (d)
Solution:
(d) Velocity, $v=\frac{d s}{d t}=3 t^{2}-12 t+3$
Acceleration, $a=\frac{d v}{d t}=6 t-12$; For $a=0$, we have, $0=6 t-12$ or $t=2 s$. Hence, at $t=2 s$ the velocity will be
$v=3 \times 2^{2}-12 \times 2+3=-9 ms^{-1}$
