Solving Differential Problem: Constant Power, Mass Acceleration

  • Thread starter Thread starter Cate
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 5K views
Cate
Messages
29
Reaction score
0

Homework Statement



An automobile with mass m accelerates starting from rest, while the engine supplies constant power P.

a) Show that the speed is given as a function of time by v= (2Pt/m)^(1/2)

b) Show that the acceleration is given as a function of time by a= (P/2mt)^(1/2)

c) Show that the displacement is given as a function of time by x-x0= (8P/9m)^(1/2)t^(3/2)


The Attempt at a Solution



I know that this isn't a difficult question but I always seem to get confused. For a) I started trying to find the derivative v= (1/2)(2P/m) ^ (-1/2)...but I don;t know how to contiue by showing that it's a function of time. If someone could just help me with a I'm sure i'll be fine for b) and c)

Thanks!
 
Physics news on Phys.org
You're trying to start the problem from the thing you're supposed to show. It isn't going to work.

One of the handful of expressions for power is P=Fv. You know that P is constant, which means that F and v must be functions of time. Since you want only one dependent variable (namely, v), here is your first subtask: Eliminate the variable F by expressing it in terms of mass and velocity. You should be able to do this using Newton's second law. Do you see what I mean?
 
I don't really understand what you mean, so Newton's second law ( F= Ma) In the place of P (for part a))Iinput FV then in the place of F I input Ma so it looks like V= (2ma(t)/m)^ (1/2) so the m's cancel out right? then what do I do?
 
Cate said:
I don't really understand what you mean, so Newton's second law ( F= Ma) In the place of P (for part a))

No, you're going to put it in place of F. And what is a, in terms of v?

so it looks like V= (2ma(t)/m)^ (1/2) so the m's cancel out right?

You won't get off that easy. You will end up with dv/dt in your equation, and you will have to integrate it to find v(t).