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The law of motion of a body moving along a straight line is $x=\frac{1}{2}$ vt. $x$ being its distance from a fixed point on the line at time $t$ and $v$ is its velocity there, Then
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acceleration $f$ is constant
We have, $\quad x=\frac{1}{2} v t$ $\Rightarrow \quad x=\frac{1}{2} \frac{d x}{d t} t$
$\left[\because v=\frac{d x}{d t}\right]$
$\Rightarrow \quad \frac{2 d t}{t}=\frac{d x}{x}$
$\Rightarrow \quad 2 \cdot \int \frac{d t}{t}=\int \frac{d x}{x}$
$\Rightarrow \quad 2 \log |t|+\log |c|=\log |x|$
$\Rightarrow \quad \log \left(t^{2} \cdot c\right)=\log x$
$\Rightarrow \quad x=t^{2} c$
$\Rightarrow$
$\frac{d x}{d t}=2 t c$
$\Rightarrow$
$\frac{d^{2} x}{d t^{2}}=2$
$\Rightarrow$ acceleration $f$ is constant.
$\left[\because v=\frac{d x}{d t}\right]$
$\Rightarrow \quad \frac{2 d t}{t}=\frac{d x}{x}$
$\Rightarrow \quad 2 \cdot \int \frac{d t}{t}=\int \frac{d x}{x}$
$\Rightarrow \quad 2 \log |t|+\log |c|=\log |x|$
$\Rightarrow \quad \log \left(t^{2} \cdot c\right)=\log x$
$\Rightarrow \quad x=t^{2} c$
$\Rightarrow$
$\frac{d x}{d t}=2 t c$
$\Rightarrow$
$\frac{d^{2} x}{d t^{2}}=2$
$\Rightarrow$ acceleration $f$ is constant.
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