# Homework Help: Velocity and acceleration of a ball

1. Jul 20, 2010

### emma402

Consider a ball whose velocity is as follows:
3·cm/s at t=0·s
19·cm/s at t=1·s,
31·cm/s at t=2·s,
39·cm/s at t=3·s, and
43·cm/s at t=4·s.

(a) Which of the following statements describes the velocity and acceleration of the ball? Choose one answer only.

1.increasing velocity, increasing acceleration
2.increasing velocity, constant acceleration
3.increasing velocity, decreasing acceleration
4.decreasing velocity, increasing acceleration
5.decreasing velocity, constant acceleration
6.decreasing velocity, decreasing acceleration

Wouldn't the answer be #5?

(b)In order to move as described in the previous part, the ball could (choose one answer)
1. roll UP a STEEP STRAIGHT SLOPE 2. roll DOWN a STEEP STRAIGHT SLOPE
3.roll UP a CURVED IN SLOPE (similar looking to a valley) 4.roll DOWN a CURVED IN SLOPE
5.roll UP a HILL CURVED OUT 6.roll DOWN a HILL CURVED OUT

I was thinking the answer would be between 1 and 2 for (b)....would it be 1 since the velocity speeds up then slows down, almost like it is going up slope?

2. Jul 20, 2010

### 6Stang7

Why would 5 be the answer for a?

3. Jul 20, 2010

### emma402

b/c as the time goes on the distance is getting shorter and shorter between seconds, so it is slowing down or decreasing is what I thought.

4. Jul 20, 2010

### 6Stang7

How is the distance getting shorter? Are you doing this:

19·cm/s at t=1·s===>19(cm/s)*(1s)=19cm

?

Your problem gives you a velocity at a given time; distance isn't listed anywhere.

Look at the values for velocity that are given. How do they change with time?

5. Jul 20, 2010

### emma402

no I was finding the difference between cm with the given seconds. I was not given any direction on how to do this kind of problem. If I do it the way you said the values for velocity keep increasing....19cm, 62cm, 117cm, 172cm...so that would make (a) #2 correct? For part (b) then I believe it would also be number #2?

6. Jul 20, 2010

### collinsmark

Message contents deleted.

Sorry, I totally misinterpreted the problem. I originally misread the velocities as position. Please ignore my previous comments, and I apologize for the confusion.

Last edited: Jul 20, 2010
7. Jul 20, 2010

### emma402

acceleration? But what numbers do I use where?

8. Jul 20, 2010

### 6Stang7

OK, maybe I am reading the problem incorrectly here, but I see it as stating that:

At t=0 seconds, the ball is moving at 3cm/s
At t=1 seconds, the ball is moving at 19 cm/s
At t=2 seconds, the ball is moving at 31 cm/s

So how is the velocity of the ball decreasing?

9. Jul 20, 2010

### emma402

if the velocity is decreasing, and the distance is increasing with time, then the acceleration is increasing? and if this is right..... then the ball would be going up hill (decrease in velocity) and increase in accelerating the opposite way (answer 5 for part (b))?

10. Jul 20, 2010

### emma402

Stang, b.c there was a difference of 16cm, then 12cm, then 8cm...the distance is getting shorter

11. Jul 20, 2010

### 6Stang7

You stated at the start:
Yes, the rate at which the velocity changes is going down, but not the velocity ;)

12. Jul 20, 2010

### emma402

ok sorry i am getting confused with whos telling me what....ok so what I have straight is that the velocity is INCREASING....now how do I work on the next part regarding acceleration? acc=change in velocity / change in time

so 62-19=43...43cm/1s....62 is from taking 31cm x 2s
117-62=55...55cm/1s
172-117=55cm/1s...is this anywhere close?

13. Jul 20, 2010

### collinsmark

My bad. Sorry about that. When I originally glanced at the problem, I incorrectly thought that the numbers were measurements of position at given times. I now realize I was mistaken. In fact, they are measurements of velocity at given times. Again sorry about that.

Allow me to start over.

Consider a ball whose velocity is as follows:
3·cm/s at t=0·s
.......In between these points, the velocity changes by 16 cm/s, per second.
.......(19 - 3 [cm/s])/(1 - 0 ) = 16 cm/s2

19·cm/s at t=1·s,
.......In between these points, the velocity changes by 12 cm/s, per second.
.......(31 - 19 [cm/s])/(2 - 1 ) = 12 cm/s2

31·cm/s at t=2·s,
.......In between these points, the velocity changes by 8 cm/s, per second.
.......(39 - 31 [cm/s])/(3 - 2 ) = 8 cm/s2

39·cm/s at t=3·s, and
.......In between these points, the velocity changes by 4 cm/s, per second.
.......(43 - 39 [cm/s])/(4 - 3 ) = 4 cm/s2

43·cm/s at t=4·s.

With that, what do the above changes in maroon represent (i.e. what does change in velocity per unit time represent)?

14. Jul 20, 2010

### emma402

this shows that the velocity is decreasing every 4cm/s..correct?

i.e. what does change in velocity per unit time represent)?...acceleration which is constant, correct?

15. Jul 20, 2010

### collinsmark

Not quite. (Again, forget what I said in my previous post before I deleted it. I was misinterpreting the problem -- And again, sorry about that).

In the first second, the velocity increases by a whopping 16 cm/s within that second (and the velocity is increasing, not decreasing).

But later, between the 3rd and 4th second, the velocity only increases by 4 cm/s within that particular second (but note that the velocity is still increasing, at least a little).

Yes, that's right!
Sorry, not so right.
If the acceleration were constant, the change in velocity per unit time would also be constant. But as you can see here, it is not constant. In the first second, the change in velocity is 16 cm/s, per second; but later on the change is only 4 cm/s, per second.

16. Jul 20, 2010

### emma402

so the accleration is decreasing and the velocity is increasing?

17. Jul 20, 2010

### 6Stang7

bingo :D

18. Jul 21, 2010

### emma402

yeah! ok so could you help me with the direction it is moving on a hill, or which hill for that matter. if the acceleration is decreasing, then will it be going up hill since the velocity is increasing the opp. way? (the choices are in the original problem) it makes me think it is the sloped in valley looking hill with the ball going up.

19. Jul 21, 2010

### 6Stang7

OK, so we know that the velocity is increasing. We also know that the value of the acceleration is decreasing. Let's work through this problem step by step.

IF the velocity is increasing, is the ball going down a hill (of some type) or up a hill (of some type)?

20. Jul 21, 2010

### emma402

it is going down a hill....