**Theif and officer motion** **A thief runs away from a police station with a uniform speed of 100m/minute. After a minute an officer starts chasing him. He runs at 100 m/minute is the first minute and increases speed by 10m/min every minute. After how many minutes does the thief get caught? ** *The answer I get is 2sqrt(5) minutes but the answer is 5 minutes.* *I assumed the thief gets caught in N minutes which is the same time the officer runs for. *
Thief ran for N+1 minutes.
therefore s=100N+100. ----------eq(1)
for the officer
u=100 , a=10m/minutes^2 , t=N
By s=ut+(at^2)(1/2)
s=100N+5N^2 ---------eq(2)
distance covered has to be same so
eq(1)=eq(2)
N=2sqrt(5) *BUT, I found out that if you form an A.P. with the distances covered by the officer in n minutes, and find the total distance covered instead of using the formula, And proceed normally, you can get N=5*
Some thing I must be missing.......please help?
*
Last edited by NK48; Jan 4th 2015 at 05:17 AM.
* |