A thief after commiting a theft …

According to question we are given that,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.

Let the time taken by police to catch the thief be n minutes.
Since, the thief ran 2 minutes before the police started running.
Time taken by thief before he was caught = (n + 2) mins
Distance travelled by the thief in (n + 2) mins = 50(n + 2) m
Given that speed of the police increased by 5 m/min.
Speed of police in the 1st min = 60 m/min
Speed of police in the 2nd min = 65 m/min
Speed of police in the 3rd min = 70 m/min
Now, this forms an A.P. with a = 60 and d = 5
{tex}\therefore{/tex} Total distance travelled by the police in n minutes
{tex}= \frac { n } { 2 } [ 2 \times 60 + ( n - 1 ) \times 5 ]{/tex}
{tex}= \frac { n } { 2 } [ 120 + 5 n - 5 ]{/tex}
{tex}= \frac { n } { 2 } [ 115 + 5 n ]{/tex}
{tex}= \frac { 5 n } { 2 } [ 23 + n ]{/tex}
After the thief was caught by the police,
Distance travelled by the thief = Distance travelled by the police
{tex}\Rightarrow 50 ( n + 2 ) = \frac { 5 n } { 2 } [ 23 + n ]{/tex}}
{tex}\Rightarrow 10 ( n + 2 ) = \frac { n } { 2 } [ 23 + n ]{/tex}
{tex}\Rightarrow{/tex}20n + 40 = 23 + n²
{tex}\Rightarrow{/tex}n² + 8n - 5n - 40 = 0
{tex}\Rightarrow{/tex}n(n + 8) - 5(n + 8) = 0
{tex}\Rightarrow{/tex}(n + 8)(n - 5) = 0
{tex}\Rightarrow{/tex}n + 8 = 0 or n - 5 = 0
{tex}\Rightarrow{/tex}n = -8 or n = 5
Thus,time is always postive, n=5
Thus, the time taken by police to catch the thief is 5 minutes.