A vehicle moving at a speed of 50 kmph has stopped in 4.5 seconds after applying the brakes to 60%. The braking distance for the same speed is

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Solution

Case I:
$V = u + a t$
$\Rightarrow 0 = (50 \times \frac{5}{18}) + (0.6 a) \times 4.5$
$\Rightarrow a = 5.144 m / s e c^{2} (retardation)$

Case II:
$V^{2} - u^{2} = 2 a s$
$\Rightarrow S = \frac{u^{2}}{2 a}$
$= \frac{{(50 \times \frac{5}{18})}^{2}}{2 \times 5.144} = 18.75 m$

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A vehicle moving at a speed of 50 kmph has stopped in 4.5 seconds after applying the brakes to 60%. The braking distance for the same speed is

Case I: V=u+at ⇒ 0=(50×518)+(0.6a)×4.5 ⇒ a=5.144 m/sec2 (retardation) Case II: V2−u2=2as ⇒ S=u22a =(50×518)22×5.144=18.75 m

Case I:
$V = u + a t$
$\Rightarrow 0 = (50 \times \frac{5}{18}) + (0.6 a) \times 4.5$
$\Rightarrow a = 5.144 m / s e c^{2} (retardation)$

Case II:
$V^{2} - u^{2} = 2 a s$
$\Rightarrow S = \frac{u^{2}}{2 a}$
$= \frac{{(50 \times \frac{5}{18})}^{2}}{2 \times 5.144} = 18.75 m$