Motion in a Straight Line - Result Question 13
13. 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: 13. (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 $=12 m / s^{2}$
