Question

The displacement ${}^{\prime }{x}^{\prime }$ (in $kg$) moving in one dimension under the action of a force, is related to time ${}^{\prime }{t}^{\prime }$ (in $sec$) by $t=\sqrt{x}+3$. The displacement of the particle when its velocity is zero will be :

• A. $2m$
• B. $4m$
• C. $0m$ (zero)
• D. $6m$

Answer 1

The correct option is C $0m$ (zero)

$t=\sqrt{x}+3$

or $\sqrt{x}=t-3$ ... (i)

Squaring on both sides of equation (i),

$x=(t-3{)}^2$ ... (ii)

$=t^2+9-6t$ 2

If v be the velocity of the particle,

Then $v=\frac{dx}{dt}=\frac{d}{dt}\left(x\right)$

$=\frac{d}{dt}(t^2-6t+9)$

$=2t-6$ 1

when $v=0,2t-6=0$

or $t=3sec$. ...(iii) 1

$\therefore$ From equations (ii) and (iii),

$x=3^2+9-6\times 3$

$=18-18$

$x=0$ 1