Velocity and acceleration of a ball

[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?

 

[6Stang7]

Why would 5 be the answer for a?

 

[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.

 

[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?

 

[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?

 

[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.

 

[emma402]

acceleration? But what numbers do I use where?

 

[6Stang7]

[Edit: I interpret that you are trying to say that the change in position per unit time is getting shorter. If you mean total distance per unit time, that's another story, and you might want to rethink the problem.]

Okay, so far so good. You have determined that the velocity is decreasing.

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?

 

[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))?

 

[emma402]

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

 

[6Stang7]

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

You stated at the start:

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.

Yes, the rate at which the velocity changes is going down, but not the velocity ;)

 

[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?

 

[collinsmark]

ok sorry i am getting confused with whos telling me what...

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.

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?

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)?

 

[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?

 

[collinsmark]

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

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).

i.e. what does change in velocity per unit time represent)?...acceleration

Yes, that's right!

which is constant, correct?

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.

 

[emma402]

so the accleration is decreasing and the velocity is increasing?

 

[6Stang7]

bingo :D

 

[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.

 

[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)?

 

[emma402]

it is going down a hill...

 

[6Stang7]

OK, so we know the ball must be going down a hill, so any answer with "up a hill" can be eliminated. Now we need to figure out what the shape of the hill is (straight, curved in, curved out).

Next, what do we know about the acceleration? It's in the same direction as the ball and the value of the acceleration is decreasing.

In the picture below, which ball is going have the higher acceleration?

 

[emma402]

the one on the left...i take it you are leading up to the point of the shapes of the hills?

 

[6Stang7]

Correct. So, the stepper the slope, the higher the acceleration value is. We know that the acceleration value starts out high, and then decreases in value over time. Since the acceleration is decreasing, what does that mean about the slope?

 

[emma402]

that it is a large slope or hill soooo it would be the hill curving out?

 

[6Stang7]

Very close; I think you might be confused by what they mean by curved out and curved in.

Below is curved in(left) and curved out(right). Which one do you think is the right answer for this problem?

 

[emma402]

sorry, I was looking at it from curve compared to the atmosphere...the one on the left would have the decreasing acceleration?

 

[6Stang7]

Correct, so the answer for b is...

 

[emma402]

just so there's no more confusion, the answer for b is the previous graph on the left...meaning the graph that curves IN when looking from the graph out b.c the graph on the left has a less steep slope than the one on the right.
am I correct?

 

[6Stang7]

Yea. Since the acceleration decreases in value with time, that means the slope must decrease in value as well. The slot of the curved in starts out really high, then progresses to low.