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Acceleration from Kinetic energy.

  • Thread starter Bryn_7
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  • #1
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Hi,
The question reads;

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!
 

Answers and Replies

  • #2
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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.
 
  • #3
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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!
 
  • #4
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Bump! Still stuck??
 
  • #5
Doc Al
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Your idea of taking a derivative is good, but what will you take the derivative of? What's the definition of power?
 
  • #6
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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?
 
  • #7
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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
 
  • #8
Doc Al
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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.
 
  • #9
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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?
 
  • #10
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Where p is Power, M mass and a acceleration?
 
  • #11
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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.
 
  • #12
Doc Al
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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.
 
  • #13
Curious3141
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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.
 
  • #14
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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!
 
  • #15
Doc Al
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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.
 
  • #16
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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!
 

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