A car is moving with speed of 150 km/h and after applying the break it will move 27 m before it stops. If the same car is moving with a speed of one-third the reported speed, then it will stop after travelling ______ m distance.

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answer is 3.

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Detailed Solution

Given, s = 27m, v = 0

$\begin{array}{l} u = 150 k m / h = 150 X \frac{5}{18} = \frac{125}{3} m / s \\ U \sin g v^{2} - u^{2} = 2 a s, w e g e t \\ 0 - \frac{125^{2}}{9} = 2 X a X 27 \Rightarrow a = \frac{- 125^{2}}{2 X 27 X 9} m / s^{2} \end{array}$

Now, when car is moving with one – time of its initial speed, we have

$\begin{array}{l} v = 0, u = \frac{1}{3} X 150 k m / h \\ u = 50 X \frac{5}{18} = \frac{125}{9} m / s \\ n o w a g a i n u \sin g v^{2} - u^{2} = 2 a s, w e g e t \\ 0 - \frac{125^{2}}{9^{2}} = 2 X (\frac{- 125^{2}}{2 X 27 X 9}) X s \\ \Rightarrow s = \frac{125^{2}}{9^{2}} X \frac{27 X 9}{125^{2}} \Rightarrow s = 3 m \end{array}$