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) $12 ms^{-2}$

Show Answer

Answer:

Correct Answer: 13. (d)

Solution:

  1. (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 $=12 m / s^{2}$