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A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to
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$t$
Let in time $t$, the body will acquire a velocity of $v$ from rest.
As, power $(P)=$ rate of doing work $=\frac{d W}{d t}$
$=\mathbf{F} \cdot \frac{d \mathbf{x}}{d t}=\mathbf{F} \cdot \mathbf{v}$
For motion in one dimension,
$P=F v \cos 0^{\circ}=F v...(i)$
From first equation of motion,
$\begin{aligned}
&v=u+a t \\
&v=a t ...(ii) (\because u=0)
\end{aligned}$
From Eqs. (i) and (ii), we get
$\begin{aligned}
P &=F \times a t=m a \times a t \quad(\because F=m a) \\
&=m a^{2} t
\end{aligned}$
As $m a^{2}$ is constant.
So, $\quad P \propto t$
As, power $(P)=$ rate of doing work $=\frac{d W}{d t}$
$=\mathbf{F} \cdot \frac{d \mathbf{x}}{d t}=\mathbf{F} \cdot \mathbf{v}$
For motion in one dimension,
$P=F v \cos 0^{\circ}=F v...(i)$
From first equation of motion,
$\begin{aligned}
&v=u+a t \\
&v=a t ...(ii) (\because u=0)
\end{aligned}$
From Eqs. (i) and (ii), we get
$\begin{aligned}
P &=F \times a t=m a \times a t \quad(\because F=m a) \\
&=m a^{2} t
\end{aligned}$
As $m a^{2}$ is constant.
So, $\quad P \propto t$
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