Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The displacement $'x'$ (in meter) of a particle of mass $'m'$ (in $kg$) moving in one dimension under the action of a force, is related to time $'t'$ (in sec) by $t = \sqrt x +3$. The displacement of the particle when its velocity is zero, will be

AMUAMU 2014Motion in a Straight Line

Solution:

Given : $t=\sqrt{x}+3$
or $\sqrt{x}=t-3$
Squaring both sides, we get
$x=(t-3)^{2}$
Velocity, $v=\frac{d x}{d t}=\frac{d}{d t}(t-3)^{2}=2(t-3)$
Velocity of the particle becomes zero, when
$2(t-3)=0$
or $t=3 s$
At $t=3 s$
$x=(3-3)^{2}=0\, m$