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# When a train is stopped by applying break it stops after travelling a distance of $50 \mathrm{m}$. If speed of train is doubled and same retarding force is applied then it stops after travelling a distance of:(A) 50 m(B) 100 m(C) 200 m(D) 400 m

Last updated date: 29th May 2024
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Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.

We can use the equivalent formula $\mathrm{d}=\mathrm{rt}$ which means distance equals rate times time. To solve for speed or rate use the formula for speed, $\mathrm{s}=\mathrm{d} / \mathrm{t}$ which means speed equals distance divided by time. To solve for time, use the formula for time, $\mathrm{t}=\mathrm{d} / \mathrm{s}$ which means time equals distance divided by speed. The average speed of an object is defined as the distance travelled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.
$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{1} \quad$ or $\quad 0=\mathrm{u}^{2}-2 \mathrm{as}_{1}$
$\mathrm{s}_{1}=\dfrac{\mathrm{u}^{2}}{2 \mathrm{a}}$
when speed of train is doubled let train travel ${{s}_{2}}$ distance
so $\mathrm{s}_{2}=\dfrac{(2 \mathrm{u})^{2}}{2 \mathrm{a}} \ldots(\mathrm{ii})$
$\left\{\right.$ given $\left.\mathrm{s}_{1}=50 \mathrm{m}\right\}$
$\mathrm{s}_{2}=4 \dfrac{\mathrm{u}^{2}}{2 \mathrm{a}} \Rightarrow \mathrm{s}_{2}=4 \mathrm{s}_{1}=4 \times 50=200 \mathrm{m}$