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+5t^{2}=100t+200

⇒5t^{2}+15t−200=0

⇒t^{2}+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.