In my textbook I have the following problem:

A ball rolls along a table then falls off the edge landing on soft sand. Sketch the s–t, v–t, and a–t graphs for its vertical motion.

The textbook's solution is the following: enter image description here There is an instant where the acceleration and the velocity become positive while still moving downwards. I don't understand how this is possible. In my sketch, I had acceleration as constant (9.81) throughout the fall until the ball reaches the ground. I don't understand how the acceleration could be positive if it is going in the negative direction.

How can I make sense of this solution?


1 Answer 1


The velocity never becomes positive during the entire duration. The velocity only decreases in the negative direction (so it goes up, towards the x-axis). You can see the trace for velocity below or at the x-axis during the entire time.

As you probably deduced, the acceleration is a negative constant during the first part (during the free fall of the ball, where net force is negative is due to gravity). However, the sudden positive acceleration during the second part is because the ball strikes the sand, which causes it to slow down. Slowing down in the downward/negative direction can be thought of as speeding up towards the upward/positive direction. Therefore, by definition, the acceleration is positive during this part (but careful again, velocity is not positive, otherwise the ball would be moving back up).

Also, the velocity doesn't go back to 0 immediately after hitting the sand because the sand is soft. It moves out of the way a little bit when the ball strikes it, so the ball will continue to penetrate the sand even after it first strikes it. But eventually, the ball will stop (which is why the acceleration and velocity at the end is 0).

Your confusion on how the acceleration can be positive even if the ball is moving in the negative direction is a common one. I'll try to give you an example to help you understand that acceleration and velocity can be in opposite directions:

Suppose you throw a ball upwards. I think we can both agree that it's velocity is positive right after leaving your hand. Now, think of the forces acting on the ball throughout its "flight". Ignoring air resistance, gravity is the only force, and it's acting in the negative direction (downwards). Because the net force and acceleration are always in the same direction, we can deduce that the ball is accelerating downwards, even though its initial velocity is upwards (which is why the ball will begin to slow down immediately).

  • $\begingroup$ but then why is the positive acceleration so high. If it slows down from -9.81 to 0, shouldn't the positive acceleration be equal in magnitude to the negative acceleration, ie 9.81? Instead the positive acceleration seems to be at least twice as large in magnitude. $\endgroup$
    – Adam Grey
    Commented Jun 21, 2019 at 21:58
  • $\begingroup$ @Adam Grey Let's think of acceleration as simply Δv/Δt. Also, let's assume that the maximum velocity reached by the ball is -10 m/s. Because the free fall period is clearly longer than the slow-down period, let's assign them more fake numbers. 10s of free fall, and 2s of slow-down. So acceleration during free fall is -10/10s = 1. But acceleration during the slowdown is +10/2s = 5. Intuitively, because the ball has the same change in velocity (10) in both cases, but because this change occurs over a much shorter period during its slow-down, the acceleration during the slow-down must be higher. $\endgroup$
    – F16Falcon
    Commented Jun 21, 2019 at 22:03
  • $\begingroup$ Sorry, I didn't see your edit. I understand your example, but my confusion is mainly on why the acceleration changed from negative to positive while falling. Your explanation of it slowing down in the negative direction, therefore being positive makes sense, but I am still confused about the magnitude of the acceleration. $\endgroup$
    – Adam Grey
    Commented Jun 21, 2019 at 22:04
  • $\begingroup$ @Adam Grey No worries. Check the comment reply I just posted and let me know if you would like for me to elaborate more. $\endgroup$
    – F16Falcon
    Commented Jun 21, 2019 at 22:05
  • 1
    $\begingroup$ Yes, I understand it now. Thank you for making it so clear (I can't upvote you yet). $\endgroup$
    – Adam Grey
    Commented Jun 21, 2019 at 22:08

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