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
B
10 m/s2
C
15 m/s2
D
18 m/s2

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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