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Question

(d)
$\sqrt{x}=t+7$
$\Rightarrow x=(t+7)^2$
$=t^2+49+14 t 	ext { (squaring) }$
$\frac{d x}{d t}=2 t+14$
$v=2 t+14 \Rightarrow v \propto t$
Accel

Vedclass · Accepted Answer

The particle moves with constant acceleration

Vedclass · Answer

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

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

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