A car is moving along a straight line is brought to a stop within a distance of $200 \mathrm{~m}$ and in…

A car is moving along a straight line is brought to a stop within a distance of $200 \mathrm{~m}$ and in time 10s. The initial speed of the car is
  1. $25 \mathrm{~ms}^{-1}$
  2. $50 \mathrm{~ms}^{-1}$
  3. $25 \mathrm{~ms}^{-1}$
  4. $25 \mathrm{~ms}^{-1}$

Solution

Given, displacement of car, $s=200 \mathrm{~m}$ Time taken, $t=10 \mathrm{~s}$ Final velocity of car $=0$ Now using, Average velocity $\times$ time $=$ Displacement or $ \left(\frac{v+u}{2}\right) \times t=s $ We get, $\quad\left(\frac{0+u}{2}\right) \times 10=200$ $ \Rightarrow \quad u=\frac{200 \times 2}{10}=40 \mathrm{~m} / \mathrm{s} $ Hence, initial speed of car $=40 \mathrm{~m} / \mathrm{s}$

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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