# Homework Help: Velocity and acceleration problem.

1. Dec 17, 2013

### alingy1

Look at the attached picture. The solutions says that only b is the right answer... Why isn't c and d good?
After all, the acceleration points in the opposite direction... The author of this book seems to reason in terms of the length of the acceleration vector. However, the acceleration vector and velocity vector have different scales so they are not comparable...

2. Dec 17, 2013

### alingy1

Here is the image.

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3. Dec 17, 2013

### nasu

The edge is not visible. And maybe you can put it upright next time. Not everyone uses a tablet. :)
Is the velocity arrow (top figure) starting at point 1?

4. Dec 17, 2013

### alingy1

Yes it is. Sorry for the confusion. I did not have access to a computer. :(

5. Dec 17, 2013

### nasu

OK, I see. It's good you question. And you are half right.
Yes, a body starting with v1 and undergoing acceleration as in the figure may have later on any of the three choices, a, b and c.

If you follow what happens, the velocity decreases continuously, while pointing to the left. At some point it will be zero and then will change direction, to the right. It will move back through the same positions in reverse. So the point with zero velocity is the farthest from A.
In order to have in point 2 these velocities shown in C and D, the particle should go to a point farther to the left, past 2 (let say 3) where it stops. Then it goes back and it passes through 2 with the velocities shown in the figure. But this will be the second time it goes through point 2.

Now, they say it moves from A to B. This is more likely to mean that they are talking about it going straight from 1 to 2 and not that is passes 2 and comes back, right?

6. Dec 18, 2013

### alingy1

Hmm... There is something going on here. Is the velocity vector that we want to find here the velocity the car has at point b? Or is it the average velocity between point A and point B? In this last case, I do not see why C and D are wrong. But, in the first case, I do understand.

7. Dec 18, 2013

### nasu

Either way. Both instantaneous velocity at point 2 and the average between 1 and 2 will point to the left.
But in the text there is nothing about an average. They say the velocity as it moves away from 2.
I understand this as instantaneous velocity.

If you draw the velocities at several points between 1 and 2, they will be like this:

2 <---- <------ <-------- <---------- <---------------- 1

As they all point to the left, the average cannot point to the right.

The only way to have the velocity pointing as in c or d is something like this:

Stop <- <-- 2 <---- <------ <-------- <---------- <---------------- 1
Ante then

Stop -> --> --->2 -----> ----------> etc.

Stop is the point where it stops, on the right hand side of 2.

8. Dec 18, 2013

### alingy1

I understand your point of view. :) Thanks for helping so far.
But, suppose the velocity was the average one:

2 (vector 0) <-- <---- <------ <-------- <---------- <---------------- 1
2-> --> ---> -----> ---------->1

C and D could work?

(Sorry for sounding really stupid. I'm teaching myself physics this winter and I'm having some trouble with some issues sometimes.)

9. Dec 18, 2013

### nasu

If you have something like this, the velocity at point 2 is zero.

You can have average velocity over an interval but not at a point. Average velocity at point 2 does not make sense.
Average velocity from 1 to 2 will be some arrow to the left.
Average velocity from 2 to 1 will be some arrow to the right.
And average velocity for the round trip will be zero.
None of these fit what the problem asks.

10. Dec 18, 2013

### alingy1

Hmm... The author of my textbook (Knight, Physics for scientists and engineers, p.14) says that the vectors connecting each dot in the motion diagram is the average velocity vectors!!! I always thought that motion diagrams showed instant velocities! This continues my confusion :P which is why I am thinking why the author did not put C as a good answer since it fits "Average velocity from 2 to 1 will be some arrow to the right."

11. Dec 18, 2013

### nasu

I don't have the book and I don't know what kind of "motion diagram" is that.