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A thief is driving away on a straight road in jeep moving with a speed of 9 m/s. A police man chases him on a motorcycle moving at a speed of 10 m/s. If the instantaneous separation of the jeep from the motorcycle is 100 m, how long will it take for the police to catch the thief?

I have solved the question in this way :

The net displacement of thief is +100 and of the police is +0.a=0.

100=10t (S=ut)

0 = 9t

So I subtracted both the equations and got correct answer t = 100 seconds. But, I don’t understand why do I subtract them?

Instead, if I add then I get 100 = 19t

I was very happy when I got this but don't understand why I did it that way.

Please help me understand this.

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  • $\begingroup$ you should include units in your work. This problem does not involve acceleration, only relative velocity between thief and police . T seconds = 100 meters/(police meters/second - thief meters/second). You got lost by not using units. $\endgroup$
    – Robert DiGiovanni
    Commented Dec 7, 2020 at 15:05
  • $\begingroup$ Thank you all for giving your time.I have understood it now. $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 15:37
  • $\begingroup$ I have answered my question $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 15:44
  • $\begingroup$ Your first equation says that the cop has reached the original position of the cyclist after ten seconds. The second says that the cyclist has reached the origin of the coordinate system (90 m down the road) in the same 10 seconds. I can't think of any logical reason for adding or subtracting these two. $\endgroup$
    – R.W. Bird
    Commented Dec 8, 2020 at 21:28
  • $\begingroup$ @R.W.Bird I have answered my question. $\endgroup$
    – user280997
    Commented Dec 9, 2020 at 9:58

4 Answers 4

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I think I am understanding it a bit now.

It is more of making the distances equal actually.

I did 10t -9t = 100 -0 because let us say if we make a graph of police at origin and theif at 100m =x.

10 t - 9t is the distance.When we get this equal , means they are at same point.

But we need another value because their is some difference in their i.e 100 distance.

10t-9t = 100-0 means It is the time when both of them will meet each other.

In other words , it is 10t = 9t +100 THIS IS THE MAIN POINT.

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  • $\begingroup$ Good job. One may opine 9t=0 only if the thief is at zero and the police is 100 meters away. But, you got it. (A follow up would be how far they both traveled before the chase ended?) $\endgroup$
    – Robert DiGiovanni
    Commented Dec 7, 2020 at 16:59
  • $\begingroup$ Oh yeah. Thank you.@RobertDiGiovanni $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 17:08
  • $\begingroup$ Great job 👏 . . $\endgroup$
    – Srijan
    Commented Dec 14, 2020 at 14:01
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When you subtracted, What you did here was you found the separation between the thief and the policeman in the LHS, and in the RHS you found the relative speed between them, as you only subtracted the velocities since t is common. Thus it is the alternate method TKA suggested.

When you add the equations we get 100=19t which is the time taken by something with a speed 19m/s to travel 100m which does not really help solve the question.

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The right way to go about it:

  • use the relative speed between the thief and the policeman which is 10m/s - 9m/s = 1m/s
  • Distance is 100m
  • use the distance time formula s = ut (without any acceleration).
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  • $\begingroup$ OP has asked for explanation of their two methods, not an alternative method. $\endgroup$
    – Daddy Kropotkin
    Commented Dec 7, 2020 at 14:12
  • $\begingroup$ @TKA Also thank you for your method so solving it. $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 14:18
  • $\begingroup$ But my question is different $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 14:18
  • $\begingroup$ Thought the approach used resulted in right answer it is not the right way to look at the problem. The initial distance between the thief and the police is given. As police moves towards the thief the thief is also moving. When police catches the thief the distance between them will be zero. But both police and thief displacement from their initial position will be something else. $\endgroup$
    – TKA
    Commented Dec 7, 2020 at 14:26
  • $\begingroup$ But I actually used equations of motion.There must be some reason for this right? $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 14:36
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Alternative approach: Define an x axis,and apply the position equation, x = $x_o$ + vt to each of the vehicles.

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  • $\begingroup$ OP has asked for explanation of their two methods, not an alternative method. $\endgroup$
    – Daddy Kropotkin
    Commented Dec 7, 2020 at 14:12
  • $\begingroup$ @N.Steinle Can you please help me here ? $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 14:16
  • $\begingroup$ @Can you help with question as well. $\endgroup$
    – user280997
    Commented Dec 7, 2020 at 14:34