Physics ut onramps

Question

This question is typical on some driver's license exams: A car moving at 44 km/h skids

15 m with locked brakes.

How far will the car skid with locked brakes at 110 km/h? Assume that energy loss is due only to sliding friction.

Answer in units of m.

Answer 1

Let

v₁ = 44 km/h be the speed in the first case,

v₂ = 110 km/h be the speed in the second case,

s₁ = 15 m be the breaking distance in the first case,

s₂ (to be calculated) be the breaking distance in the second case.

m (unknown be the mass of the truck),

g be gravitational acceleration,

f (unknown) be the coefficient of friction.

In general, if the truck is moving at speed v, then it has kinetic energy mv²/2.

This energy must be dissipated by friction, i.e., must equal the work of friction.

The force of friction is mgf, and its work after distance s is mgfs.

Thus

mv²/2 = mgfs

v²/2 = gfs

In our two cases

v₁²/2 = gfs₁

v₂²/2 = gfs2

Divide the second equation by the first

s₂/s₁ = (v₂/v₁)²

s₂ = (v₂/v₁)²s₁

Substituting actual numbers

s2 = (110/44)²×15 = 93.75 m