# Acceleration as a function of displacement

2
1. The problem statement, all variables and given/known data

All right, the problem is that I have a motor that's acceleration at a rate of a = 6 theta rad/s^2. For the bigger problem that this is part of, I need to find out the velocity function. How do I do this if a isn't given in terms of t?

2. Relevant equations

omega = d theta/dt a = d omega / dt

3. The attempt at a solution

All I can think of is that

a = 6 theta
integrate both sides for t

v = 6 theta t
theta = 3 theta t^2? that doesn't make sense!

and theta is really theta(t), so how can I integrate that function if I don't know what i is?

help me calculus gurus!

2. ### Gib Z

3,348
I Think you have to use the result $$a=x''=\frac{d}{dx} (\frac{1}{2}v^2)$$.

That result can be obtained as such - $$\frac{d}{dx} ((1/2) v^2) = \frac{d}{dv} ((1/2) v^2) \cdot \frac{dv}{dx}=v\frac{dv}{dx}$$ by the chain rule. Another application of the chain rule: $$v\frac{dv}{dx} = \frac{dx}{dt}\cdot\frac{dv}{dx} = \frac{dv}{dt}=a$$

2
actually as it turned out:

a(t) = 6 theta

better put as

a(t) = 6 theta(t)

a(t) is just the second derative of theta(t), so

theta''(t) = 6 theta(t)

tossing away the 6 mentally for a second, what function equals its own second derative? e does! so let's play with that and see what we get

if theta(t) = e^t, then theta''(t) = e^t. Okay, but we need to bring that 6 back in. So, theta (t) = e^sqrt6 t, theta'(t) = v(t) = sqrt6 e^sqrt6 t, and v''(t) = a(t) = 6 e^sqrt6 t.

To check it:a(t) = 6 e^sqrt6 t. e^sqrt6 t = theta(t) substitute and we get back to the original a(t) = 6 theta.

yay :) using this I got the right answer for my problem!

But I'm on another problem (last one, promise) with similar premises.

omega(t) = 5 (theta(t))^2 is what I'm given.

So theta'(t) = 5 (theta(t))^2

Toss away the 5, we can deal with it later...but what function equals its derative when you square it?