# Acceleration from Kinetic energy.

Hi,

A car motor produces 20Kw of power. All of which is transfered to the wheels producing drive, there is no resistance. the mass is 800kg. Calculate the acceleration after 7 seconds.

I can see that it requires using dv/dt = a and rearanging Ke=1/2 mv^2 to include acceleration but when I tried using. v=u+at to form Ke= 0.5 M (AT)^2 I got an incorrect answer.

HELP!

v=u+at

This is not right because acceleration is not constant. Do you remember what the definition of "power" is? Start from that: assuming that power and mass are constants, you should be able to solve the speed of the car as a function of time.

This is not right because acceleration is not constant. Do you remember what the definition of "power" is? Start from that: assuming that power and mass are constants, you should be able to solve the speed of the car as a function of time.

I knew it was wrong because of that, it was getting the velocity as a function of time which I was struggling with ?

Thanks for looking!

Bump! Still stuck??

Doc Al
Mentor
Your idea of taking a derivative is good, but what will you take the derivative of? What's the definition of power?

Well I thought that power is rate of change of Ke in this circumstance, so d(Ke)/dt , but after that I got stuck. I thought if I could differentiate velocity with respect to time it may help?

Well I thought that power is rate of change of Ke in this circumstance, so d(Ke)/dt , but after that I got stuck. I thought if I could differentiate velocity with respect to time it may help?

So if P = dKe/dt, then Ke = ∫Pdt. Solve this for velocity

Doc Al
Mentor
Well I thought that power is rate of change of Ke in this circumstance, so d(Ke)/dt , but after that I got stuck.
That's the right idea. Show what you got when you took the derivative.
I thought if I could differentiate velocity with respect to time it may help?
Well, sure. dv/dt = a, which is what you are trying to solve for.

so If I differentiate both sides with respect to time:

d(Ke)/dt = 0.5M(V^2) dv/dt

I would get P = MA ?

But that is incorrect as A varies?

Where p is Power, M mass and a acceleration?

so If I differentiate both sides with respect to time:

d(Ke)/dt = 0.5M(V^2) dv/dt

That's not right. Remember chain rule! Now you have v=v(t), and mass is constant. What is d/dt v(t)2?

I would get P = MA ?

But that is incorrect as A varies?
I dont understand where you got that from.

Doc Al
Mentor
so If I differentiate both sides with respect to time:

d(Ke)/dt = 0.5M(V^2) dv/dt
That's not quite right. Remember the chain rule.

Curious3141
Homework Helper
There is no need to use calculus here. An application of Power = Force * velocity and another of F = ma will allow expression of Power in terms of mass, (instantaneous) acceleration and (instantaneous) velocity. Express v in terms of the other variables.

Then use ½mv^2 = KE = power*time. Put the previously derived expression for v here and solve for a in terms of power, mass and time.

That's not quite right. Remember the chain rule.

So using the chain rule would you treat V like a trig function? Then when differentiating with respect to time would you get,

P = MAV ?

Where P is d(Ke)/dt so Power

M is the Mass which is constant

A is the Acceleration

V is the velocity.

That gets the correct answer but is it the correct way? If it is thank you all so much for the help. It seems really simple now!

Doc Al
Mentor
So using the chain rule would you treat V like a trig function?
You'd treat v as a function of t. d(KE)/dt = d(KE)/dv*dv/dt.

Then when differentiating with respect to time would you get,

P = MAV ?
Exactly.

You'd treat v as a function of t. d(KE)/dt = d(KE)/dv*dv/dt.

Exactly.

Thank you so much! I'm not quite sure why I had the block on treating V as a function. It seems so simple now!