Question

If the displacement of a particle varies with time as $\sqrt{x}=t+7$, then

• A. Velocity of the particle is inversely proportional to $t$
• B. Velocity of the particle is proportional to $t^{2}$
• C. Velocity of the particle is proportional to $\sqrt{t}$
• D. The particle moves with constant acceleration

Answer 1

Answer: (D)

$\sqrt{x}=t+7$
$\Rightarrow x=(t+7)^{2}$
$=t^{2}+49+14 t$ (squaring)
$\frac{d x}{d t}=2 t+14$
$v=2 t+14 $
$\Rightarrow v \propto t $
Acceleration:
$a=\frac{d v}{d t}$
$a=2\, ms ^{-2} \rightarrow$ constant