Physics Help Forum Theif and officer motion

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 Jan 4th 2015, 04:15 AM #1 Junior Member   Join Date: Feb 2014 Posts: 6 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 04:17 AM.
 Jan 5th 2015, 07:08 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,333 Your approach is correct only if you can assume a constant value for acceleration, which implies a linear equation for velocity (i.e. v = 100 m/s + 10m/s^2 x t). But the police officer's velocity is not linear at all, but rather is a step function. Hence the equation s = ut + (1/2)at^2 does not apply. Instead you have to calculate the distance between the officer and thief for each minute. At t=1 minute the gap is 100 meters, and at t = 2 the gap is reduced by 10 meters since from t = 1 to t=2 the officer runs at 10 m/minute faster than the thief. So at t=2 the gap is 90 m. From t=2 to t=3 the officer runs 20 m/min faster than the thief, so the gap closes by another 20 meters, and the distance between them is now 90-20 = 70 meters.. Continue in like fashion and you find that the gap is closed at t=5 seconds.
Jan 8th 2015, 08:31 AM   #3
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Join Date: Feb 2014
Posts: 6
 Originally Posted by ChipB Your approach is correct only if You can assume a constant value for acceleration.......
but isn't 10m/m^2 a constant value? he changes a evry minute by the same value so......... Just askin my doubts

Jan 8th 2015, 08:51 AM   #4
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,333
 Originally Posted by NK48 but isn't 10m/m^2 a constant value? he changes a evry minute by the same value so......... Just askin my doubts
Here are plots of the officer's velocity and acceleration. Note that the velocity graph has step-wise changes in it, whereas if the acceleration was constant the velocity graph would be a straight line. And since acceleration is the slope of velocity you can see that it is spiky - not constant at all.
Attached Thumbnails

 Jan 14th 2015, 10:25 AM #5 Junior Member   Join Date: Feb 2014 Posts: 6 wow that just made it a lot more clear..thanks for all your help Chip

 Tags motion, officer, theif