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# Acceleration-Time Graph watch

1. Today my physics teacher brought up an old dispute he and I had about an exam question that we looked at months ago, I'm looking here to clarify the answer. The question was asking us to draw an acceleration-time graph of a bouncing ball, ignoring the acceleration at the moment of collision and ignoring resistive forces (Downwards is the positive direction). My initial thoughts were that acceleration due to gravity is constant, therefore the graph is a straight line at +9.81m/s^2 for entire time. When we came to look over the question however my teacher argued that the graph should first be a straight line at +9.81m/s^2 and then a vertical line down and another straight line at -9.81m/s^2. I cannot see how the acceleration would become negative when the ball is returning based on my teachers explanation, despite the ball slowing down, the change in velocity is still positive i.e for a perfectly efficient bounce, when falling u=0 and v=10, the change in velocity is +10 and when it is returning upwards, u=-10 and v=0, the change in velocity is also +10 as velocity is initially negative and the time taken is exactly the same to return up, therefore the acceleration is constant. So can someone confirm my answer or prove my teachers answer. Thanks
2. Hi there! Thanks for posting the doubt!
* the ball accelerates with a constant amount of 9.81m/s^2
* as ball bounces up, it has negative acceleration hence -9.81
*as ball moves down, it has positive acceleration , however on the graph it will be on the negative side
because anything moving downward is considered as negative on at graph

So it should be a straight line of -9.81m/s^2 in my opinion. Let's see what others have got to say on it.
http://www.vivaxsolutions.com/maths/altravelgraphs.aspx

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Updated: March 9, 2018
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