Question

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

Answer 1

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

Answer 2

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.

Answer 3

So would it start off at 2m/s?

Answer 4

Yes

Answer 5

Now what is the velocity at t = 1?

Answer 6

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

Answer 7

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

Answer 8

yep

Answer 9

t=2 would be 0

Answer 10

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

Answer 11

It goes from 2 m/s2 to ?

Answer 12

1

Answer 13

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?

Answer 14

5

Answer 15

yep

Answer 16

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

Answer 17

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

Answer 18

yes the acceleration is 0.

Answer 19

so that means the velocity doesn't change.

Answer 20

Right

Answer 21

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

Answer 22

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

Answer 23

So at t = 4, v = ?

Answer 24

why not 2?

Answer 25

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

Answer 26

why not subtract 2?

Answer 27

Because the acceleration is -3 right not -2?

Answer 28

Sorry. Was getting confused with the +2

Answer 29

ok so what is the velocity at t = 4

Answer 30

v=2

Answer 31

yep

Answer 32

t = 5?

Answer 33

the acceleration is back to 0 again right?

Answer 34

v=2

Answer 35

yep

Answer 36

So i think thats it. Any questions?

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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.

Answer 44

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

Answer 45

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

Answer 46

Ok well I guess its just an estimate then.

Answer 47

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

Answer 48

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?

Answer 49

Err.. nvm

Answer 50

I do not.

Answer 51

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)

Answer 52

-3

Answer 53

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

Answer 54

and a is the acceleration

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

Right. It's -3

Answer 59

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

Answer 60

4.87

Answer 61

*4.88

Answer 62

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

Answer 63

Yes

Answer 64

Are you doing this:

\[(-3m/s^{2})*(3.41 s - 3 s)+5 m/s = v_{f}\]

\[(-3)(0.41)+5=v_{f}\]

\[-1.23 + 5 = v_{f}\]

Answer 65

I see what I did now.

Answer 66

Great

Answer 67

I got 3.77 the second time I did it.

Answer 68

yep thats what i got

Answer 69

Thanks for helping out.

Answer 70

NP