The displacement ' $x$ ' (in meter) of a particle of mass ' $m$ ' (in kg ) moving in one dimension u

Question

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 $t=\sqrt{x-2}+4$. The displacement of the particle when its velocity is zero, will be

Professor PC Thomas Classes · Answer

(a) : Given $t=\sqrt{x-2}+4$ or $\sqrt{x-2}=t-4$ \ Squaring both sides, we get $x-2=(t-4)^{2} \implies x = (t-4)^{2} + 2$ \ Velocity, $v=\frac{d x}{d t}=\frac{d}{d t}\left[(t-4)^{2} + 2ight]=2(t-4)$ \ Velocity of the particle becomes zero, when $2(t-4)=0 \implies t=4 \mathrm{~s}$. \ At $t=4 \mathrm{~s}, x=(4-4)^{2} + 2 = 2 \mathrm{~m}$