The velocity (v) of a particle starting from rest increases linearly withtime $(t)$ as $v=4 t$, where $v$ is…
The velocity (v) of a particle starting from rest increases linearly withtime $(t)$ as $v=4 t$, where $v$ is in $m s^{-1}$ and $t$ is in second. The distance covered by the particle in the first 4 seconds is
$16 \mathrm{~m}$
$32 \mathrm{~m}$
$8 \mathrm{~m}$
$64 \mathrm{~m}$
Solution
The velocity of the particle is given by
$
\mathrm{v}=4 \mathrm{t}, \mathrm{u}=0
$
From the equation of motion
$
\begin{aligned}
& \mathrm{v}=\mathrm{u}+\mathrm{at} \\
& 4 \mathrm{t}=\mathrm{at} \Rightarrow \mathrm{a}=4 \mathrm{~m} / \mathrm{s}^2 \\
& \mathrm{~S}=4 \mathrm{t}+\frac{1}{2} \mathrm{at}^2=0 \times \mathrm{t}+\frac{1}{2} \times 4 \times 4^2 \\
& \mathrm{~S}=32 \mathrm{~m}
\end{aligned}
$