# Homework Help: Finding change in speed from x vs t graph

1. Jun 17, 2014

1. The problem statement, all variables and given/known data

For a particle moving in one dimension, x vs t graph is given below.

At each point state whether speed is increasing, decreasing or not changing.

2. Relevant equations

Just the basic relation between position, velocity (first derivative of position) and acceleration (second derivative of position).
Also, when this graph is concave up (curved upwards) the acceleration is positive. And if concave down (curved downwards) the acceleration is negative.
When v and a have same sign the speed is increasing; if v and a have opposite sign speed is decreasing.

3. The attempt at a solution

For point P:
Speed is not changing because there is no curvature at that point (ie acceleration is zero)

For point Q:
The curve is concave down so a<0 and from the graph the slope of tangent line at Q is zero so v=0

For point R:
Speed is not changing because there is no curvature at that point (ie acceleration is zero)

For point S:
The curve is concave up so a>0 and from the graph the slope of tangent line at S is zero so v=0

Now I have no problem with point P and R. But I am not sure what to say about point Q and S.

Unfortunately, to add to my confusion, the answer in the book says the speed in decreasing for both Q and S.

By the way the problem is from Young & Freedman, University Physics 12e

2. Jun 17, 2014

### Orodruin

Staff Emeritus
Note that there is a difference between speed and velocity. Velocity has a direction, speed does not.

3. Jun 17, 2014

### dauto

I would say the speed is not changing in any of the points

4. Jun 17, 2014

Here is the answer the book gives: (which doesn't clear up anything for me)
At point Q: v>0 and a<0 so speed decreasing
At point S: v<0 and a>0 so speed decreasing

How on earth they came to this conclusion???

5. Jun 17, 2014

### HallsofIvy

To the left of Q the x is increasing so the speed is positive. To the right of Q, x is decreasing so the speed is negative. The speed is going from positive to negative so it is decreasing and the acceleration is negative.

The situation is the opposite at S.

6. Jun 17, 2014

Ok I understand.

So shouldn't it be that the speed is increasing as the particle passes S? (negative to zero to positive).
Whereas the book says, speed is decreasing in both cases. Is there a mistake in the answer key?

Last edited: Jun 17, 2014
7. Jun 17, 2014

But wait...

Speed is the magnitude of (instantaneous) velocity. As far as I know magnitude of a vector can not be negative. Please explain.

Last edited: Jun 17, 2014
8. Jun 18, 2014

Speeding up or slowing down? (1D kinematics question)

[EDIT: This was the start of a new thread. But for being similar, it is duly merged here with this one by PF moderator.]

We know when a particle's velocity and acceleration are in same direction (ie have same signs) the particle is speeding up. And when velocity and acceleration are in opposite directions (ie have different signs) the particle is slowing down.
It is clear that when acceleration is zero, the particle is neither speeding up nor slowing down.

But what if, v = 0 and a > 0 ?
Similarly, what if, v = 0 and a < 0 ?
In theses two cases, is the particle speeding up or slowing down?

Last edited: Jun 18, 2014
9. Jun 18, 2014

### Matterwave

Both cases the speed is increasing, but the particles are moving in different directions.

When a particle has a velocity v and an acceleration with opposite sign, at some time, the particle will stop. If that time comes and the acceleration is still "on", then the particle will start accelerating in the other direction.

Accelerations also don't have to be parallel to the velocity. You can apply an acceleration to a particle perpendicular to the direction of its motion. This will tend to produce circular motion. So, the correct way to think about things is from a vector perspective. An acceleration will tend to change the velocity in the direction that it is applied. In fact, acceleration is defined as the change of velocity (also a vector, so the direction matters) per unit time.

10. Jun 18, 2014

### sophiecentaur

The same rules apply for all signs an values of v and a n straight line motion. You just have to include the fact that -v is velocity in a backwards direction. -a, when applied for long enough, will result in a negative v.
Google SUVAT equations. These equations will give answers for all motion in a straight line and constant acceleration values.

11. Jun 18, 2014

Thanks I understand.

But still I have some confusions regarding this topic.

Last edited: Jun 18, 2014
12. Jun 18, 2014

### jbriggs444

In mathematical usage, neither would hold. The rate of speeding up has a discontinuity when v=0, going from -a to +a (or from +a to -a). Said a different way, the one-sided first derivative of speed is positive on the future side and negative on the past side. A physicist would likely avoid the problem by talking about velocity rather than speed.

In ordinary English usage, one could say that it is speeding up either way. Customary usage is prejudiced to talking about future behavior and not so much about past behavior. An English speaker would often avoid the issue by using a different phrase entirely and say that the particle is "starting to move".

13. Jun 18, 2014

Thanks everyone. But I really need a final clear-up. So I shall restate my problem more clearly again:

Below is the x vs t graph for a particle moving in 1D. Now when the particle is at Q and later at S, is it speeding up or slowing down?

Most of the people are saying that the particle is speeding up. Which makes sense to me. Because at point P and Q velocity is zero. So when there is acceleration (no matter in which direction) it's speed will increase.

So far so good. My main confusion arises when I check out the answer from the book. The book says:
at point Q: v > 0, a < 0, speed decreasing
at point S: v < 0, a > 0, speed decreasing

Now someone please put me out of this uncomfortable situation by saying the answer of the book is wrong (and if not so, then why?).

For reference: The question is from Young and Freedman "University Physics 12e" Chapter 2: test your understanding question 2.3(d) page 47. The answer is given at page 61.

14. Jun 18, 2014

### jbriggs444

If v > 0 and a < 0 then velocity is becoming less positive -- approaching zero from above. Speed is equal to the magnitude of velocity and is decreasing.

If v < 0 and a > 0 then velocity is becoming less negative -- approaching zero from below. Speed is equal to the magnitude of velocity and is decreasing.

15. Jun 18, 2014

I agree about the acceleration. But why are they saying v > 0 at point Q? And why it would be v < 0 at point S? Isn't that inconsistent with the given graph? What I see is v = 0 at both Q and S.

Last edited: Jun 18, 2014
16. Jun 18, 2014

### ehild

You are right, the velocity is zero both at Q and S. The velocity is the time derivative of the position/time (x(t)) graph. The velocity is positive before reaching Q, and its magnitude decreases with time. The acceleration is negative. The speed is also decreasing. At Q, the point has zero velocity and turns back. It accelerates backward, the acceleration is negative, but the speed is increasing.

It is just the opposite at S. The point moves backwards, so x(t) decreases, the velocity is negative, but its magnitude decreases, so the acceleration is positive. At S, the velocity is zero, and the point turns back again, so its velocity becomes positive. The acceleration is positive. Before S, the speed decreases, and it increases after S.

ehild

17. Jun 18, 2014

### jbriggs444

Apologies for the abbreviated response in post #14 above. I did not have access to the drawing to see that the velocity was in fact zero at Q and S.

I think that we all agree that before reaching Q, speed is decreasing and that after leaving Q, speed is increasing. The question seems to be about what happens exactly at Q.

The speed is neigher increasing nor decreasing there. If you look at the speed versus time graph at Q it will look a lot like the graph of the absolute value function at zero. The slope is negative prior to Q, positive after Q and undefined at Q.

Similarly at S. If the book's answer actually says that speed is increasing at Q and at S then the book is wrong.

Is that what you wanted to hear?

18. Jun 18, 2014

### CWatters

The slope is positive just before Q and negative just after Q. So it must be zero somewhere in between. I think it's fair to assume they choose point Q to be the point where it is zero.

19. Jun 18, 2014

### jbriggs444

The mean value theorem applies for continuous functions. In an ideal one-dimensional situation, there is no reason to think that the first derivative of speed is continuous.

In the situation at hand, the rate of change of speed changes discontinuously. While that may be unphysical, it is what the model requires.

20. Jun 18, 2014

That is quite interesting for me. So you are saying that actually x-t graph for 1D should not be as smooth as it is drawn at points like Q and S?
That means, x(t) is not differentiable at Q?

In other words -
if I draw a v vs t graph from this x vs t graph there should be a discontinuity at point Q (and S) in the v vs t graph...
... Did I understand you correctly?

21. Jun 18, 2014

### ehild

According to the original problem, x(t) is the position in case of an one-dimensional motion. v(t) is the derivative of x(t). The function looks smooth, differentiable, so v(t) is continuous. It is not sure that v(t) is also smooth, but in case of most Physical problems, it is also differentiable, the acceleration is also a continuous function.

As an example, the height of a vertically projected stone can be written as x(t) = vot-gt2/2. The velocity is v= dx/dt=vo-gt. It is positive for t<vo/g, negative for t>vo/g and zero at t=vo/g. The acceleration is dv/dt=-g.
The maximum of the function is at t=vo/g, and the maximum value is x(max)=vo2/(2g).

The speed is the magnitude of the velocity. speed=|v| It is not smooth.

But the acceleration is the time derivative of the velocity.

ehild

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22. Jun 18, 2014

### ehild

There is no discontinuity in he velocity v(t) at the extrema of the x(t) function. Even the speed is not discontinuous.

ehild

23. Jun 18, 2014

### jbriggs444

Right.

Position graphed over time is continuous
Velocity graphed over time is continuous
Acceleration graphed over over time is continuous.

You can keep on taking derivatives forever and the result will still be (for essentially all physical situations modeled classically) continuous. That is because the equations of classical mechanics are second order partial differential equations.

Speed graphed over time is continuous. That is because speed is the absolute value of velocity and "absolute value" is a continuous function.

But the rate of change of speed graphed over time is not continuous. That's because although "absolute value" is continuous, it is not everywhere differentiable.

24. Jun 18, 2014

Sorry that I misunderstood at first. (mixed up speed and velocity... Oops!)

So finally I think I can say for the graph in question -
Speed in decreasing before Q
Speed is increasing after Q
Speed is neither increasing nor decreasing at Q

Thanks a lot to everyone for your patience in explaining!

Unfortunately, if you good people never mind, I have one last (perhaps dumb) question.
What does neither increasing nor decreasing means here? Constant or Not defined?

25. Jun 18, 2014

Not defined.