A thief is driving a motorcycle at 36 km/h on a straight road.

Question

A thief is driving a motorcycle at 36 km/h on a straight road. He crosses a point where a police car is waiting. The moment when the thief crosses the police car, the police car starts from rest and increases its speed at a uniform acceleration of 1 m/sec2 . Eventually the police car catches up with the thief. After how many seconds the police car just overtakes the thief?

Answer 1

Let

v = 36 km/h = 36000m/3600s = 10 m/s be the speed of the thief,

a = 1 m/s² be the police car acceleration,

s1(t) be the distance of the thief from the intersection after time t,

s2(t) be the distance of the police car from the intersection after time t.

We want to calculate t such that s1(t) = s2(t).

Write the equations for s1 and s2:

s1(t) = vt (because the thief has constant speed)

s2(t) = at²/2 (because the police car has constant acceleration starting from rest)

Set s1(t) = s2(t)

vt = at²/2

Express t as

t = 2v/a

Substitute actual numbers

t = 2×10/1 = 20 s.

Note how ridiculously slow is the police chase.