Question

39. A car speeding at 30km/h is stopped in 8m by applying brakes. If the same car were travelling at 60km/h, it can be brought rest with the same braking power in: a) 32 m b) 18m c) 16m d) 2m​

Answer 1

Answer:

Explanation:

first let me derive the formula for you.

let the initial velocity be u, and the distance covered be s.with constant retardation a brought to rest,so final velocity=0

so v²=u²+2as

as a is retardation,it is negative

so 0²=u²-2as

u²=2as⇒ s=u²/2a

here,the same car is being considered,so the value of a will be same in both cases.

so if a is constant,then we can write s ∝ u²

so we can also write s₁/s₂=u²₁/u²₂ (hope u understand this,if we take direct proportionality symbol,when taking the ratio,it will come like tis

ok then let me explain, s ∝ u² means we can write s=ku²,where k is a constant,so if we take another s and u values with a being constant,for simple simplification,we take the ratio s₁/s₂=u²₁/u²₂,that means we can grt rid of that constant k which we gave,as it will get cancelled in numerator and denominator

so here assuming s₁=8m, u₁=30km/h ,u₂=60km/h, we have to find s2 which is our required answer.

so s₁/s₂=u²₁/u²₂= 8/s₂=30²/60²

s₂=(8×60×60)/(30×30)= 8×2×2=32m

answer -opt a)

Hope this helps you my dear friend

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thanks :)