# 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? - Algebra

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+(t1)×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+15t200=0

t2+3t40=0

(t+8)(t5)=0

So, t = −8 or t = 5

∴ t = 5  (As t cannot be negative)

Thus, the policeman catches the thief after 5 min.

Is there an error in this question or solution?