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The graph below is meant to represent an a-t graph for the following situation: A point moves away from origin at a slow constant velocity for 5s Then it moves away at a medium-fast constant velocity for 5s Then it stands still for 5s Then it moves toward the origin at a slow constant velocity for 5s Finally it stands still for 5s

enter image description here

My problem is that I am getting the last two bars as being of equal length. This is because in the third to last stage, it is simply standing still. Therefore, the acceleration it takes to move toward the origin at a slow constant velocity must be equal and opposite to the one it requires to go back to rest. This just seems to be common sense to me although it can be proved as follows.

When it is at rest, v=0. Let the acceleration required to move again be -a (-ve since towards origin), so because this is instantaneous (since the velocity is constant after), the constant velocity we get at the last stretch is 0-a, or just -a. Finally, to go back from this velocity to 0, we need an acceleration of +a. So the two bars at the end should be equal, one is -a the other is +a.

Raghib
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1 Answers1

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I obtain this time-graph for acceleration and velocity:

enter image description here

user
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  • Shouldn't the bar at t=10 for acceleration be bigger than at t=5 (because at t=5 the car already had a velocity, so if it is to go to rest at t=10 then the deceleration at t=10 must be larger than the acceleration at t=5). – Raghib Jan 01 '18 at 11:15
  • "A point moves away from origin at a slow constant velocity for 5s" say V, "Then it moves away at a medium-fast constant velocity for 5s" say V/2 "Then it stands still for 5s" V=0. Thus the acceleration in bith case decrease the velocity of V/2. – user Jan 01 '18 at 11:21
  • Apologies if I am not understanding this, however I still think I am correct. Here is what I thought: it moves at V at first. Then at t=5, it is still at V. However, at t=5, it begins moving at medium-fast, which is faster than V (since V was "slow"). So after t=5 it is moving lets say at 1.5V. Then at t=10, it comes to rest, so it must have an instant deceleration of 1.5V. So the bar at t=5 is 0.5V (since it goes from V to 1.5V) but at t=10 it is -1.5V, which is bigger. – Raghib Jan 01 '18 at 11:24
  • Oh yesof course I've interpreted the velocity description in a different way! I revise the graph! Sorry – user Jan 01 '18 at 11:26
  • @Raghib I've just updated the graph! – user Jan 01 '18 at 11:50