Car Work, Acceleration and Power

[yb1013]

Homework Statement

As the driver steps on the gas pedal, a car of mass 1100 kg accelerates from rest. During the first few seconds of motion, the car's acceleration increases with time according to the expression below.

a = (1.160 m/s^3)t - (0.210 m/s^4)t^2 + (0.240 m/s^5)t^3

(a) What work is done on the car by the wheels during the interval from t = 0 to t = 3.00 s?

____________ J

(b) What is the wheels' output power at the instant t = 3.00 s?

___________ W

The Attempt at a Solution

Im really quite confused about this one... Although it seems like a problem where derivatives might be involved to simplify this a little more. Am I on the right track at all? Please Help??

 

[LowlyPion]

How do you determine work? What is the formula?

 

[yb1013]

well W = P x t

 

[LowlyPion]

What about Force? Can't that give you the work?

 

[yb1013]

yea, W = F x d

 

[LowlyPion]

yea, W = F x d

There you go.

Now the car is accelerating isn't it? So it's applying a force to the car, the F = m*a kind of thing.

But it also does it over a distance covered in 3 sec.

So if there was only a way to figure the distance covered in 3 sec that might prove useful?

 

[yb1013]

i might be totally going the wrong route here but to get the distance part, can you just like take two anti-derivatives to bring it back to position which would give you distance??

also for the F=ma, how do you know what the acceleration is? can you just plug 0 and 3 into the equation and take the average of that?

 

[LowlyPion]

i might be totally going the wrong route here but to get the distance part, can you just like take two anti-derivatives to bring it back to position which would give you distance??

also for the F=ma, how do you know what the acceleration is? can you just plug 0 and 3 into the equation and take the average of that?

You're given the acceleration, so yes it looks like you can just integrate twice to determine x position distance at 3. You know initial condition is at rest.

Edit: Forgot you were trying to get to Work, so you can integrate once then for the Velocity function. Your work then should be equal to the kinetic energy at t=3 right?

 

[yb1013]

hmm, i did all that but it didnt come out right..

 

[LowlyPion]

Forgot you were trying to get to Work, so you can integrate once then for the Velocity function. Your work then should be equal to the kinetic energy at t=3 right?

 

[yb1013]

hmm maybe, alright well thanks for you help!