Function of Time Homework: Solve for A, B, and C

[veronicak5678]

Homework Statement

A bird's flight is given by the following function:
v = (At^2 - B) i + (Ct) j

1)If the bird is traveling due north at 3 m/s at t= 6 seconds, find 'C'.
2) The bird accelerates at (-2 m/s^2)i + (.5 m/s^2)j at t = 4s. Find 'A' and 'B'.

Homework Equations

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The Attempt at a Solution

I really don't know what to do. If I just solve part 1 algebraically, I get .5 m/s^2. That can't be all there is to it, because then how do I solve part 2?

 

[olgranpappy]

Homework Statement

A bird's flight is given by the following function:
v = (At^2 - B) i + (Ct) j

1)If the bird is traveling due north at 3 m/s at t= 6 seconds, find 'C'.
2) The bird accelerates at (-2 m/s^2)i + (.5 m/s^2)j at t = 4s. Find 'A' and 'B'.

Homework Equations

?

The Attempt at a Solution

I really don't know what to do. If I just solve part 1 algebraically, I get .5 m/s^2.

what do you mean "solve part 1 algebraically"? what exactly did you do? show the work and it will be easier for us to help you.

That can't be all there is to it, because then how do I solve part 2?

 

[olgranpappy]

...for example, you probably used C*6=3... but what else does "due north" tell you?

 

[veronicak5678]

Sorry about that. You were right. All I did was set Ct = 3.00 m/s

C ( 6 seconds) = 3 m/s

C = (3 m/s) / (6 s)

Do I need to break this into vectors? Like (3sin 90) i ? Where does t come in?

 

[olgranpappy]

Sorry about that. You were right. All I did was set Ct = 3.00 m/s

Yes, that's quite the right thing to do... but also there is something else that the statement "due north" tells you.

It tells you that at t=6 the coefficient of the 'i' term is zero.

Just like at t=6 the coefficient of the 'j' term is 3.

So, again, at t=6 the whole coefficient of the 'i' term (the thing in parenthesis next to the 'i') equals zero. Write this equation down. Once you tell us this equation we will move on to the next step.

 

[veronicak5678]

A(6^2) - b = 0

 

[olgranpappy]

A(6^2) - b = 0

Right, again. This gives us an equation for B in terms of A. B=36A.

Next we move on to exploit the information given about the acceleration.

At any time (t) we know that the x-velocity is
$$v_x=(At^2-B)$$

The x-acceleration is the derivative with respect to time of the x-velocity
...what is the value of this derivative?

 

[veronicak5678]

A= 2At

 

[olgranpappy]

Yes,
$$a_x=2At$$

...then from "2)" you know that at t=4 the x-acceleration is -2

Use this in the above equation to find A.

 

[veronicak5678]

I understand. What happens for the value of B?

 

[olgranpappy]

I understand. What happens for the value of B?

Well, now you know A... and you have an equation for B in terms of A, remember:

B=36A

 

[veronicak5678]

Oh! Of course. I think I've been sitting at the computer too long. You have been a great help. Thanks a lot!

 

[olgranpappy]

Oh! Of course. I think I've been sitting at the computer too long. You have been a great help. Thanks a lot!

You're welcome. Cheers.