http://oi58.tinypic.com/qx5lvs.jpg

- anonymous

http://oi58.tinypic.com/qx5lvs.jpg

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 4n1m0s1ty

Ok this problem is pretty much the same as the last one.

- 4n1m0s1ty

The only difference is that we are dealing with velocity and acceleration, rather than position.
I think you probably already know that the acceleration is the change in velocity over time. So, in this problem we kind of treat velocity like position in the last problem, and acceleration like velocity previously.

- anonymous

So would it start off at 2m/s?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 4n1m0s1ty

Yes

- 4n1m0s1ty

Now what is the velocity at t = 1?

- 4n1m0s1ty

The acceleration was 2 m/s2 at t = 0, so after 1 second, we give a boost to the initial 2m/s velocity.

- anonymous

So would it be 4 since we are adding 2 to the 2m/s of t=0?

- 4n1m0s1ty

yep

- anonymous

t=2 would be 0

- 4n1m0s1ty

Nope. What's the acceleration at t = 1?

- 4n1m0s1ty

It goes from 2 m/s2 to ?

- anonymous

1

- 4n1m0s1ty

yes, so our velocity is at 4 m/s, and we just add 1 m/s after 1 s, so the velocity at t =2 is?

- anonymous

5

- 4n1m0s1ty

yep

- 4n1m0s1ty

so now for t = 3. Look at the acceleration at t =2 and add is to your velocity.

- anonymous

Then would t=3 be 0 since it's a line?

- 4n1m0s1ty

yes the acceleration is 0.

- 4n1m0s1ty

so that means the velocity doesn't change.

- anonymous

Right

- anonymous

Would the be negative numbers since the acceleration is in the negative now?

- 4n1m0s1ty

so when the acceleration is negative, we are reducing the positive velocity. So at t =3, your velocity is 5 m/s, and we subtract 3 m/s now since its negative

- 4n1m0s1ty

So at t = 4, v = ?

- anonymous

why not 2?

- 4n1m0s1ty

Are you saying why not v = 2 at t = 3?

- anonymous

why not subtract 2?

- 4n1m0s1ty

Because the acceleration is -3 right not -2?

- anonymous

Sorry. Was getting confused with the +2

- 4n1m0s1ty

ok so what is the velocity at t = 4

- anonymous

v=2

- 4n1m0s1ty

yep

- 4n1m0s1ty

t = 5?

- 4n1m0s1ty

the acceleration is back to 0 again right?

- anonymous

v=2

- 4n1m0s1ty

yep

- 4n1m0s1ty

So i think thats it. Any questions?

- anonymous

Okay so 0s=2
1s=4
2s=5
3s=0
4s=2
5s=2

- anonymous

Sorry it took so long to type that. My browser kept freezing up on me.

- 4n1m0s1ty

3s = 5, the acceleration was 0, not the velocity

- anonymous

Okay so if we took the velocity at 3.41s how do we go about doing that?

- 4n1m0s1ty

look between t = 3 and t = 4 on the velocity graph you just drew.

- anonymous

Looks to be about 3.50 to me...does that seem reasonable?

- 4n1m0s1ty

Maybe, I'm not sure how your assignment system works. I'm not sure if it wants you to calculate it exactly or just by estimation. Did your instructor give you any equations to work with.

- 4n1m0s1ty

Just for my information is this a high school or college physics course?

- anonymous

He didn't. It's college but this is the first time I've ever had physics.

- 4n1m0s1ty

Ok well I guess its just an estimate then.

- anonymous

It didn't take 3.50 so I think it's wanting more exact.

- 4n1m0s1ty

Ok well, then that would require calculating the slope of the graph between t = 3 and t = 4. Do you know how to do that?

- 4n1m0s1ty

Err.. nvm

- anonymous

I do not.

- 4n1m0s1ty

The information is already given. The slope between t = 3 and t =4 is the acceleration. So whats the acceleration in that time frame equal to (look at the accel graph)

- anonymous

-3

- 4n1m0s1ty

ok so the equation for constant acceleration is this:
\[a = \frac{ v_{f} - v_{i} }{ t_{f} - t_{i} }\]
vf is the final velocity
vi is the initial velocity
tf is the final time
ti is the initial time

- 4n1m0s1ty

and a is the acceleration

- 4n1m0s1ty

so we know a is -3,
vi is 5m/s,
ti is 3s,
and tf is 3.41s
now we just need to solve for vf to find our velocity at 3.41 s

- anonymous

(2m/s-5m/s)/(4s-3s)

- 4n1m0s1ty

that would give you the acceleration, but we already found it from the graph.

- anonymous

Right. It's -3

- 4n1m0s1ty

so now we need to find the velocity at t = 3.41,
\[-3 m/s^{2} = \frac{ v_{f} - 5 m/s}{ 3.41 s - 3 s }\]
so solve for vf

- anonymous

4.87

- anonymous

*4.88

- 4n1m0s1ty

I'm getting a different answer, I would check your math. Are you using -3 for the acceleration?

- anonymous

Yes

- 4n1m0s1ty

Are you doing this:
\[(-3m/s^{2})*(3.41 s - 3 s)+5 m/s = v_{f}\]
\[(-3)(0.41)+5=v_{f}\]
\[-1.23 + 5 = v_{f}\]

- anonymous

I see what I did now.

- 4n1m0s1ty

Great

- anonymous

I got 3.77 the second time I did it.

- 4n1m0s1ty

yep thats what i got

- anonymous

Thanks for helping out.

- 4n1m0s1ty

NP

Looking for something else?

Not the answer you are looking for? Search for more explanations.