A thief runs with a uniform speed of 100 m/minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief.

#### Solution

Suppose the policeman catches the thief after *t *minutes.

Uniform speed of the thief = 100 metres/minute

∴ Distance covered by thief in (*t* + 1) minutes = 100 metres/minute × (*t *+ 1) minutes = 100 (*t* + 1) metres

Distance covered by policeman in *t* minutes = Sum of *t *terms of an AP with first term 100 and common difference 10

=`t/2`[2×100+(t−1)×10] m

=t[100+5(t−1)]

=t(5t+95)

=5t^{2}+95t

When the policeman catches the thief,

5t^{2}+95t=100(t+1)

⇒5t^{2}+95t=100t+100

⇒5t^{2}−5t−100=0

⇒t^{2}−t−20=0

⇒ *t* = − 4 and *t* = 5

∴* t *= 5 (As *t* cannot be negative)

Thus, the policeman catches the thief after 5 minutes.