# Rectilinear Motion

1. Jan 23, 2008

[SOLVED] Rectilinear Motion

1. The problem statement, all variables and given/known data A car starts from rest and moves along a straight line with acceleration $a=3s^{-1/3}m/s^2$ where s in meters. Determine the cars acceleration at t =4 seconds.

I know that I am given acceleration as a function of position. I know that by the chain rule
$$a=dv/dt=(dv/ds)(ds/dt)=\frac{dv}{ds}\cdot v$$ I am just not sure where to go from here?

Any hints?

Thanks,
Casey

2. Jan 23, 2008

### blochwave

Good start, now separate variables and use a*ds=3s^(-1/3)*ds=vdv and integrate both sides

To determine the constant use the fact that at t=0, v=0, as given in the problem and go from there

3. Jan 23, 2008

Okay so now I have $$3s^{-1/3}ds=v\cdot dv$$

$$\Rightarrow 3*\frac{3}{2}s^{2/3}+C=v+C$$

$$\Rightarrow \frac{ds}{dt}=\frac{9}{2}s^{2/3}+C$$

but, I am still in function of s not t...where am I screwing this up?

Thanks for the help so far blochwave!

4. Jan 23, 2008

### blochwave

Oh ok, so the thing here is that the way this is set up the car doesn't start at s=0

If it did, what would be its acceleration at t=0? 3 divided by the cubed root of...0? So the car's initial acceleration is ERROR, unless I'm reading that wrong

So are you perhaps given an initial value to use for s? If you were you could of course say that's what s equals when v=0 and find that constant, then separate variables again and get an s(t), find s at t=4, and put that into the equation for a

EDIT: I THINK what you can do is just go ahead and solve for that first C in terms of So or whatever, if you don't know what it is

Go ahead and separate variables again and you'll have s as some function of time including another constant. Now when t=0, s=So still, so you can solve for that constant in terms of So, and see if you can do some trickery to find So using those two equations. But I don't think that's gonna work

Last edited: Jan 23, 2008
5. Jan 23, 2008

It says "car starts from rest" which implies that at s(0)=0.

6. Jan 23, 2008

### blochwave

car starts from rest implies that if you have v(t), then v(0)=0, it says nothing about the position.

I could have a rock I drop off a cliff, it starts from rest but is its initial position 0?

7. Jan 23, 2008

My instructor e-mailed me a long hint. I can't seem to make sense of it. It says :

"use $$a=dv/dt=(dv/ds)(ds/dt)=\frac{dv}{ds}\cdot v$$

Then $a=\frac{dv}{ds}*v$ (1)

Integrate (1) to get $v=3s^{1/3}$

Use $v=\frac{ds}{dt}$ to find $s=(2t)^{3/2}$ "

My problem is the very first integration. When I integrate (1) I am getting

$$\Rightarrow \frac{ds}{dt}=\frac{9}{2}s^{2/3}+C$$ .

Not $v=3s^{1/3}$

arrgh!

8. Jan 23, 2008

I don't see why I cannot make any headway on this. I took a nap and everything! Maybe someone could shed a little light on this?

9. Jan 23, 2008

Can anybody help me out here? I really need to get this last one. Thanks!!

10. Jan 23, 2008

### blochwave

oh lordy I dunno how we screwed this up

the integral of v*dv isn't v + C, it's v^2/2

Multiply both sides by two and square root and and I bet it cleans it up to what your professor got