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)

$\sqrt{x}=t+7$
$\Rightarrow x=(t+7{)}^2$
$=t^2+49+14t$ (squaring)
$\frac{dx}{dt}=2t+14$
$v=2t+14$
$\Rightarrow v\propto t$
Acceleration:
$a=\frac{dv}{dt}$
$a=2m{s}^{-2}\to$ constant