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Question

A driver has to apply sudden brakes to stop his vehicle moving at $$72 \   {km}/{hr}$$ and stops within $$2 \ seconds$$. The retardation produced is then


A
20 m/s2
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B
10 m/s2
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C
15 m/s2
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D
18 m/s2
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Solution

The correct option is B $$10 \   {m}/{{s}^{2}}$$
Using the first equation of motion,     $$v = u + at$$
Here,  $$v = 0$$
           $$u = 72 \times \dfrac{1000}{3600} = 20 \ m/s$$
           $$t = 2s$$
Hence, $$0 = 20 + a\times 2$$
 $$a = -\displaystyle\frac{20}{2} = -10 \  {m}/{{s}^{2}}$$

Since, sign of $$a$$ is negative. Hence, retardation with magnitude $$10 \ m/s^2$$ is produced.

Physics

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