Motion In A Straight Line Ques 7
7. The motion of a particle along a straight line is described by equation : $x=8+12 t-t^{3}$ where $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero, is :
[2012]
(a) $24$ $ ms^{-2}$
(b) zero
(c) $6$ $ ms^{-2}$
(d) $12$ $ ms^{-2}$
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Answer:
Correct Answer: 7.(d)
Solution:
- (d) $x=8+12 t-t^{3}$
The final velocity of the particle will be zero, because it retarded.
$\therefore v=0+12-3 t^{2}=0$
$3 t^{2}=12$
$t=2 sec$
Now the retardation
$a=\frac{d v}{d t}=0-6 $ $t$
$a[t=2]=-12$ $ m / s^{2}$
retardation $=12 $ $m / s^{2}$
BETA
