JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 3)
The distance travelled by a particle is related to time t as $$x=4\mathrm{t}^2$$. The velocity of the particle at t=5s is :-
$$\mathrm{25~ms^{-1}}$$
$$\mathrm{20~ms^{-1}}$$
$$\mathrm{8~ms^{-1}}$$
$$\mathrm{40~ms^{-1}}$$
Explanation
$$
\begin{aligned}
& x=4 t^2 \\\\
& v=\frac{d x}{d t}=8 t
\end{aligned}
$$
At $\mathrm{t}=5 ~\mathrm{sec}$
$$ \mathrm{v}=8 \times 5=40 \mathrm{~m} / \mathrm{s} $$
At $\mathrm{t}=5 ~\mathrm{sec}$
$$ \mathrm{v}=8 \times 5=40 \mathrm{~m} / \mathrm{s} $$
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