KinematicsEasy

Braking car

A car travels at v₀ = 72 km/h and brakes with constant deceleration a = −5 m/s².\nHow many metres does it travel before stopping?

Given data

v₀ = 72 km/h = 20 m/sa = −5 m/s²v_f = 0 (stops)

Steps

0 / 3

Convert the initial velocity v₀ = 72 km/h to m/s.
Write your working — the numerical result must appear in the text m/s
How long does the car take to stop? (in seconds)
What is the braking distance in metres?

Full worked solution

Convert the initial velocity v₀ = 72 km/h to m/s.
$v_0 = 72 \div 3.6$
$v_0 = 72\,\mathrm{km/h} \div 3.6 = \mathbf{20\,m/s}$ . Converting km/h to m/s is done by dividing by 3.6 since $1\,\mathrm{km/h} = 1/3.6\,\mathrm{m/s}$ .
How long does the car take to stop? (in seconds)
$t = \dfrac{v_f - v_0}{a} = \dfrac{0 - 20}{-5}$
$t = (v_f - v_0)/a = (0 - 20)/(-5) = \mathbf{4\,s}$ . With constant deceleration $a = -5\,\mathrm{m/s^2}$ , the car takes 4 seconds to go from $20\,\mathrm{m/s}$ to rest.
What is the braking distance in metres?
$x = v_0 t + \tfrac{1}{2} a t^2 = 20 \cdot 4 + \tfrac{1}{2}(-5)(4)^2$
$x = v_0 t + \frac{1}{2}a t^2 = 20\cdot4 + \frac{1}{2}\cdot(-5)\cdot4^2 = 80 - 40 = \mathbf{40\,m}$ . Alternatively, $v_f^2 = v_0^2 + 2ax \;\rightarrow\; x = v_0^2/(2|a|) = 400/10 = 40\,\mathrm{m}$ .

Result:The car travels 40 m before stopping. Quick formula:

x = v_0^2/(2|a|) = 400/10 = 40\,\mathrm{m}

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