Average velocity given acceleration and time?

[dark-ryder341]

Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t) = bt^2 - ct^3, where b = 2.30 m/s^2 and c = 0.120 m/s^3.

(a) Find the average velocity of the car for the time interval t=0 to t=10 s

(b) Calculate the instantaneous velocity of the car at t=0, t=5 s, and t=10 s

Homework Equations

Not sure...for the first one, I thought it might be simply v = x2-x1/t2-t1 but then I realized I was given acceleration rather than distance so that won't work. Will I have to use integrations?

The Attempt at a Solution

I have no idea how to start either of these, but I have a feeling I have to integrate them and I'm really confused on that concept. =\

Thanks in advance,

- Grace

 

[kuruman]

Homework Statement

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t) = bt^2 - ct^3, where b = 2.30 m/s^2 and c = 0.120 m/s^3.

(a) Find the average velocity of the car for the time interval t=0 to t=10 s

(b) Calculate the instantaneous velocity of the car at t=0, t=5 s, and t=10 s

Homework Equations

Not sure...for the first one, I thought it might be simply v = x2-x1/t2-t1 but then I realized I was given acceleration rather than distance so that won't work. Will I have to use integrations?

What makes you say you are given acceleration?

 

[dark-ryder341]

I said that because b and c are in m/s^2 and m/s^3, not meters...unless that doesn't matter? =\ But I see what you mean, it saying distance...

 

[kuruman]

Yes, it says distance. If you look at the right side you multiply b(m/s²)*t²(s²) and you end up with meters. Same thing with c. This is dimensional analysis. I trust you know how to proceed from here.

 

[dark-ryder341]

Oh wow, I can't believe I didn't see it before. Thanks so much for your help, it makes sense now...