Question

A car traveling at 45 m/s can brake to a stop within 10m. Assume that the braking force is effective for stop and constant with speed. If the car speeds up to 90 m/s, what is its stopping distance? Provide your explanation using principle of work and energy.

Answer 1

40 m

Step-by-step explanation:

A car travelling at 45 ms has a kinetic energy of

Ec = 1/2 * m * v^2

We do not know the mass, so we can use specific kinetic energy:

ec = 1/2 * v^2

In this case

ec = 1/2 * 45^2 = 1012 J/kg

If it stops in 10 m, the braking force performed a sppecific work of 1012 J/kg in 10 m

L = F * d

F = L / d

F = 1012 / 10 = 101.2 N/kg

If the car is running at 90 m/s, the specific kinetic energy of:

ec2 = 1/2 * 90^2 = 4050 J/kg

With the same braking force the braking distance is:

d2 = L2 / F

d2 = 4050 / 101.2 = 40 m