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Homework Help: Minimum distance to avoid collision using relative motion

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data

    2 cars with velocities 10 m/s and 20 m/s are travelling in opposite directions, having uniform retardation of 2m/s^2 and 1m/s^2 respectively. Find minimum separation between them such that they don't collide.
    2. Relevant equations
    $$v^2=u^2 + 2as$$

    3. The attempt at a solution
    It can easily be solved by finding the distance they require to come to a stop using equations of motion, and adding them up, i.e. $$d_{min}=\frac{v_{1}^2}{2a_{1}} + \frac{v_{2}^2}{2a_{2}}$$. This gives, upon substituting, $$225 $$metres as the answer, which is correct.
    The problem is, I want to do it using relative motion. Suppose 1st car is moving along positive X axis towards right, and car 2 towards left. In frame of 2nd car, $$v_{12}=v_{1}-(-v_{2})=v_{1}+v_{2}=20+10=30 m/s$$. Similarly, acceleration of car 1 must be towards left (opposite to velocity) and towards right for car 2. Thus, $$a_{12}= -3m/s^2$$. Now, in frame of car 2, car 1 comes in with a velocity 30m/s and retardation (-3)m/s^2. In order to avoid collision, it has to cover the distance between them such that its velocity upon reaching 2 becomes zero(the distance here is the minimum distance), i.e. $$v_{12f}=0$$. Using $$v^2=u^2 + 2as$$, we get $$(0)^2=(30)^2+2(-3)(x)$$, which on solving gives
    I don't understand what is my mistake. Please point it out. Thanks in advance!!
  2. jcsd
  3. Oct 14, 2015 #2


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    Which car stops first? What happens to the relative acceleration then?
  4. Oct 15, 2015 #3
    @DEvens, Do you mean to say that after one of the cars stop, the relative acceleration changes? Well it does but then does it mean that the 150m I get is valid only until one of the car stops, and the other still travels 75 more metres?
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