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Here is the full question and the part that I am getting trouble with is the second part:

A block is moving from left to right. The frictional force acting on a small block of mass $0.15\ \mathrm{kg}$ has a magnitude of $0.12 \ \mathrm N$. The block is set in motion from a point X on the surface, with a speed of $3\ \mathrm{ m/s}$. It hits a vertical wall at a point Y on the surface $2$ seconds later. The block rebounds from the wall and moves directly towards X before coming to rest at the point Z. (to the right of X). At the instant that the block hits the wall, it loses 0.072J of its kinetic energy. The velocity of the block, in the direction from X to Y is v ms at time $t\ \mathrm s$ after it leaves X.

i) Find the values of $v$ when the block arrives at Y ($1.4 \ \mathrm{m/s}$) and when it leaves Y ($1\ \mathrm{m/s}$ in the opposite direction) and the value of t when the block comes to rest at Z (t=3.25 s). Sketch the velocity-time graph. (done)

ii) The displacement of the block from X in the direction from X to Y is $s\ \mathrm m$ at time $t\ \mathrm s$. Sketch the displacement-time graph of the whole journey.

So here is my problem, to sketch the displacement time graph I will use SUVAT equations of motion. I am not having problem with the first one. : it's $s=ut + 0.5at^2$. substituting $u = 3$ and $a= -0.8$. (as there is a deceleration due to friction (I have calculated it)). and that gives me $s = 3t - 0.4t^2$ , a nice n-shaped quadratic curve between the limits $t=0$ and $t=2$. But for the journey in the opposite direction : from Y to Z, I have done my calculations seperately, treating this journey as the displacement from Y to Z. But on my graph i will need to show the whole journey from X to Z. So the only SUVAT equation i'm getting for the whole journey is the first one mentioned above. As the second one ($s= t - 0.4t^2$) is not related to the first one. My question is how do i find an equation that is suitable for the second part of the journey that relates it to the first one? in my book the answer has a u-shaped curve for the second part of the journey. Why is this? Can you explain in detail?

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2 Answers 2

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The velocity starts at 3 and drops linearly until the block hits the wall. Then it become negative and heads back toward zero. The displacement (from the starting position) starts at zero and rises rapidly at first but less rapidly as the block approaches the wall. At the wall the displacement (still positive) starts to drop somewhat rapidly, but less rapidly as it approaches the end point. Before the bounce, the velocity is positive and the acceleration is negative. After the bounce, the velocity is negative and the acceleration is in the positive x direction.

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  • $\begingroup$ i guess my true question would be, why does my displacement-time graph have to look like this: ibb.co/QpvnNw7 when my formula is telling me that it should look like this: ibb.co/JR1bvV6 $\endgroup$
    – butterfly
    Oct 5, 2021 at 15:55
  • $\begingroup$ Your final equation gives the displacement from the wall (at x = 4.4 m) as the block is headed back (in the negative x direction) toward x = 0. Your first graph shows the displacement from x = 0. (Don't forget that x is on the vertical axis.) $\endgroup$
    – R.W. Bird
    Oct 6, 2021 at 14:28
  • $\begingroup$ i think you should have said y = 4.4 and 0. because y is the vertical axis on an x-y plane $\endgroup$
    – butterfly
    Oct 6, 2021 at 14:40
  • $\begingroup$ can you edit your comment? and you did not answer my question which was the WHY of it all? $\endgroup$
    – butterfly
    Oct 6, 2021 at 15:36
  • $\begingroup$ Your graphs are displacement versus time. There is no ( y). $\endgroup$
    – R.W. Bird
    Oct 7, 2021 at 13:41
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If your calculations are correct then the equation would be $$ f(x) = 3t-0.4t^2 (0<t<2) and$$ $$f(x) = 3t-0.4t^2 -((t-2)-0.4(t-2)^2) for (t>2)$$ or $$f(x) = 4.4 -((t-2)-0.4(t-2)^2)$$

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  • $\begingroup$ Note that during the duration t>2, the velocity will slowly become zero, and as friction acts to oppose the relative velocity of the 2 surfaces in contact, after the velocity becomes 0, friction will stop acting on the block and the block will become stationary $\endgroup$
    – Tim Crosby
    Oct 4, 2021 at 15:41
  • $\begingroup$ I guess my real question would be: why does the displacement-time graph look like this: ibb.co/QpvnNw7 for the part t>2 when my formula is telling me that it should look like this: ibb.co/JR1bvV6 $\endgroup$
    – butterfly
    Oct 5, 2021 at 14:28
  • $\begingroup$ i guess my true question would be, why does my displacement-time graph have to look like this: ibb.co/QpvnNw7 when my formula is telling me that it should look like this: ibb.co/JR1bvV6 $\endgroup$
    – butterfly
    Oct 5, 2021 at 14:34
  • $\begingroup$ After t>2, the block rebounds after hitting the wall, i.e its velocity and displacement are "negative ". So in any case your graph is incorrect as the displacement should decrease $\endgroup$
    – Tim Crosby
    Oct 5, 2021 at 17:51
  • $\begingroup$ yes , i know that it's incorrect because the displacement must decrease but what I truly want to find out is why should the displacement-time graph that I get from my equation be upside down? i mean how does my answer become the answer that there Is in the book , just by flipping it upside down? $\endgroup$
    – butterfly
    Oct 6, 2021 at 13:05

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