Finding n for Constant*s^n: Homework Help

[raphile]

Hey, I suspect that this is probably quite simple, but I'm a bit stuck on it, or the 2nd part at least.

Homework Statement

The displacement s of a particle moving in a straight line as a function of time t is given by s^3 = t. Find the value of n if at any time t:

constant * s^n

represents: (i) the speed of the particle; (ii) the force acting on the particle.

The Attempt at a Solution

For (i), I tried to derive a differential equation by writing:

ds/dt = k*s^n (where k is constant)
=> ds = k*t^(n/3) dt

By integrating both sides:
s = [k/((n/3)+1)] t^((n/3)+1) + c

I then hypothesised that we wanted ((n/3)+1) to be 1/3, because s=t^(1/3).
Hence, n = -2 is my answer. Is it right, or am I off-track?

For part (ii), I don't know. I know we can write force = mass * acceleration, hence F = m*s'' (s differentiated twice) but that doesn't seem to give me an equation I can solve. I know acceleration can be written in other ways, so should I write it as dv/dt or possibly v*dv/ds?

Thanks.

 

[rohanprabhu]

As for the first answer, your approach should lead you to the correct answer, but isn't. Check your solution again.

A better solution would be to write a function like:

$$s = t^\frac{1}{3}$$

and then differentiate it. The thing is, you won't have to deal with Differential Equations at all. You can differentiate it once and twice giving the answer to your questions.

 

[raphile]

Thanks, but using that method, I still get n = -2. Here's what I did:

Start with s=t^(1/3)
Differentiate: ds/dt = (1/3)t^(-2/3)

We want speed to be k*s^n and k*s^n = k*t^(n/3) because s=t^(1/3), hence:

k*t^(n/3) = (1/3)t^(-2/3)
=> So we have n/3 = -2/3 => n = -2.

I also tried differentiating a second time to get the second part as you suggested:

d2s/dt2 = (-2/9)t^(-5/3)

We want force (= mass * acceleration) to be k*s^n, hence:

k*s^n = k*t^(n/3) = (-2m/9)t^(-5/3)
=> So we have n/3 = -5/3 => n = -5.

Where am I going wrong?

 

[rohanprabhu]

As for the first question, n = -2 is the correct answer. I got confused thinking that v = kt^n rather than v = ks^n. Sorry 'bout that.

And yes, n = -5 is the right answer too.