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)
=5t2+95t
When the policeman catches the thief,
5t2+95t=100(t+1)
⇒5t2+95t=100t+100
⇒5t2−5t−100=0
⇒t2−t−20=0
⇒ t = − 4 and t = 5
∴ t = 5 (As t cannot be negative)
Thus, the policeman catches the thief after 5 minutes.