# Car Work, Acceleration and Power

#### yb1013

1. 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

3. 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??

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#### LowlyPion

Homework Helper
How do you determine work? What is the formula?

well W = P x t

#### LowlyPion

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

yea, W = F x d

#### LowlyPion

Homework Helper
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

Homework Helper
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

Homework Helper
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!

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