A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 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 = 50 m/min
∴ Distance covered by thief in (t + 2) minutes = 50 m/min × (t + 2) min = 50 (t + 2) m
The distance covered by the policeman in t minutes is in AP, with 60 and 5 as the first term and the common difference, respectively.
Now,
Distance covered by policeman in t minutes = Sum of t terms
=`t/2`[2×60+(t−1)×5]
=`t/2`[115+5t] m
When the policeman catches the thief, we have
`t/2`[115+5t]=50(t+2)
115t+5t2=100t+200
⇒5t2+15t−200=0
⇒t2+3t−40=0
⇒(t+8)(t−5)=0
So, t = −8 or t = 5
∴ t = 5 (As t cannot be negative)
Thus, the policeman catches the thief after 5 min.
