Question : 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?
Doubt by Charu
Solution :
Let the policeman catch the thief after t minute but note that the thief started running one minute before the policemen so from the thief point of view he will be caught after (t+1) minutes.
Policeman will catch the thief when the distance covered by thief will become equal to the distance covered by policeman.
Distance covered by thief in (t+1) minutes = 100 (t+1) m
Distance covered by Policeman in 1st minutes = 100 m
Distance covered by Policeman in 2nd minutes = 110 m and so on
So,
Distance covered by Policeman in t minutes
Here, a = 100, d= 10 and n = t
Sn = n/2 [2a + (n-1)d]
St = t/2[2×100 + (t-1)10]
St = t/2[200 + 10t - 10]
St = t/2[190 + 10t ]
NOW
Distance covered by thief in (t+1) minutes = Distance covered by Policeman in t minutes
100 (t+1) = t/2 [190+10t]
100 t + 100 = t [ 95 + 5t]
100 t + 100 = 95t + 5t2
5t2 + 95t - 100t - 100 = 0
5t2 -5t - 100 = 0
t2 - t - 20 = 0
t2 - (5-4)t -20 = 0
t2 -5t + 4t - 20 = 0
t(t-5) +4 (t-5) = 0
(t+4)(t-5) = 0
t + 4 = 0 & t - 5 = 0
t = -4 & t = 5
But t ≠ -4 [Time can't be negative]
∴ t = 5 minutes
Hence, the time taken by the Policeman to catch the thief = 5 minutes