When you double the speed of a car, it takes ________ times more distance to stop it.

Question

Question

This question was previously asked in

RRB ALP Electronics Mechanic 21 Jan 2019 Official Paper (Shift 1)

Answer 1

Let us suppose the initial velocity (u) and the final velocity (v) of a car. So here v = 0 the vehicle stops.

By 3^rd law of equation of motion,

v² - u² = 2aS₁

0 - u² = 2aS₁

S₁ = \(\rm \frac{-u^2}{2a}\) ...... (i)

When the initial velocity is doubled then u = 2u

v2 - u2 = 2aS₂

2aS₂ = -(2u)²

S₂ = \(\rm \frac{-4u^2}{2a}\) .......(ii)

From equation (i) and equation (ii)

\(\rm \frac{S_1}{S_2}=\left(\frac{-u^2/2a}{-4u^2/2a}\right)\)

\(\rm \frac{S_1}{S_2}=\frac{1}{4}\)

S₂ = 4S₁

Hence, when we double the speed of a car, the stopping distance increases by a factor of four.