Acceleration from Kinetic energy.

[Bryn_7]

Hi,
The question reads;

A car motor produces 20Kw of power. All of which is transferred 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!

 

[clamtrox]

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.

 

[Bryn_7]

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!

 

[Bryn_7]

Bump! Still stuck??

 

[Doc Al]

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

 

[Bryn_7]

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?

 

[clamtrox]

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]

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.

 

[Bryn_7]

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?

 

[Bryn_7]

Where p is Power, M mass and a acceleration?

 

[clamtrox]

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 don't understand where you got that from.

 

[Doc Al]

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]

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.

 

[Bryn_7]

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]

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.

 

[Bryn_7]

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!