# Basic Dynamics

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.

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.

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

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?

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.