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1) At first, the acceleration opposes the direction of velocity and the object must be slowing down.

2) At t=4, is the object’s velocity would have been at rest if the acceleration changed to 1.5 m/s^2 as the acceleration would now be in the direction of velocity and would work to increase velocity.

3) But here, the acceleration at t=4 abruptly changes to 2 not 1.5 so the velocity doesn’t have to be zero.

4) The velocity at t=7 would be the same velocity as was at t=0.

Is my interpretation of this graph correct? Is there anything else that can be determined about the nature of motion from the graph?

Edit: Point 2 is inaccurate, I think.

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    $\begingroup$ Do you know the starting conditions? From your text, I assume an object is moving in one dimension. Does it have a starting velocity? Or is it at rest at t=0? Negative acceleration does not mean that you oppose the direction of velocity since velocity can also be negative. It is just with reference to your coordinate system. $\endgroup$ – lmr Mar 26 at 7:53
  • $\begingroup$ @Imr It’s one-dimensional, yes. There’s nothing given about the initial velocity. I didn’t think of the velocity being possibly initially negative, so the only definite inference I can make about the relation between the direction of velocity and acceleration is that at t=4, the acceleration either changes to oppose velocity or being in the same direction as velocity? $\endgroup$ – Brenda Mar 26 at 8:02
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    $\begingroup$ I agree with your point 4). This can easily be seen since the area above the graph until t=4 equals the area under the graph from t=4 to t=7. To your other points, 1) depends on the starting state while 2) & 3) are unclear. $\endgroup$ – Steeven Mar 26 at 8:19
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For 1) the body could be increasing in speed in the negative direction ie if the body started from rest.

Without any further information the graph enables you to find the change in velocity which is the area under the graph.

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Since this is starting to get too long for comments, here is what you can do with the limited amount of information:

Let us assume a given starting position $x_0$ and an initial starting velocity $v_0$. Your object is moving in one dimension, let it be called the $x$-dimension. From the graph, you obtain the info about the acceleration: At first, the acceleration is negative $a_x^1 = -1.5 m/s^2$ for $\Delta t = 4s$. Then the acceleration is positive $a_x^2 = 2 m/s^2$ for $\Delta t = 4s$.

You could now compute the position $x_{t=4s}$ of your object at $t=4s$ by considering acceleration, initial position, and initial velocity:

$x_{t=4s} = \frac{1}{2} a_x^1 (\Delta t)^2 + v_0(\Delta t) + x_0$

Your velocity $v_{t=4s}$ at this point will be:

$v_{t=4s} = v_0 + a_x^1 (\Delta t)$

Subsequently, you can do the same for the second part of the motion, where the acceleration changes. You will have a different starting velocity and position though ($x_{t=4s}, v_{t=4s}$). Since this is not exactly a homework, but kind of similar, I'll stop here. Can you do the rest by yourself?

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  • $\begingroup$ I’m positive that I can. (I can’t figure how to include symbols and convenient notation, I’d comment my answer) $\endgroup$ – Brenda Mar 26 at 8:35
  • $\begingroup$ Check this Meta-Post: physics.meta.stackexchange.com/questions/804/… $\endgroup$ – lmr Mar 26 at 8:58
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I've added some images which might help. Please don't judge them quantitatively. I've added them just for an idea of what's happening. All the five velocity-time graphs have the same acceleration time graph as $a=\frac{\text{d}v}{\text{d}t}$ and that $a-t$ graph is the one you have drawn.

2) At t=4, is the object’s velocity would have been at rest if the acceleration changed to 1.5 m/s^2 as the acceleration would now be in the direction of velocity and would work to increase velocity.

The velocity at t=4 might be zero but not necessarily as you can see from the $v-t$ graphs. It might be positive or negative or zero.

4) The velocity at t=7 would be the same velocity as was at t=0.

Yes, you are correct about this point.

We have, $$v_7=v_0+(-1.5)\times(4-0) + 2\times(7-4)$$ $$\implies v_7=v_0+(-1.5)\times4 + 2\times3$$ $$\implies v_7=v_0$$

However in real life, the vehicle should take some finite time (however small that time interval may be) to change the rate of change of its velocity, acceleration, but that won't happen instantaneously.

enter image description here enter image description here enter image description here enter image description here enter image description here

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Velocity is the integral of acceleration. If we plotted the velocity-time graph, it would slope downwards at a constant gradient from t=0-4s, corresponding to the negative acceleration, and then slope upwards at a constant gradient from t=4-8s, corresponding to the positive acceleration. If we assume that at t=0s, the velocity is 0m/s, then at t=4s, the velocity is (-1.5)*(4)=-6m/s. Therefore regardless of what the next acceleration is, the velocity at t=4s is -6m/s. If the next acceleration was +1.5m/s^2, we would only achieve v=0m/s at t=8s.

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