Rectilinear Motion

  • #1
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[SOLVED] Rectilinear Motion

Homework Statement

A car starts from rest and moves along a straight line with acceleration [itex]a=3s^{-1/3}m/s^2[/itex] 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
[tex]a=dv/dt=(dv/ds)(ds/dt)=\frac{dv}{ds}\cdot v[/tex] I am just not sure where to go from here?

Any hints?

Thanks,
Casey
 

Answers and Replies

  • #2
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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
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Okay so now I have [tex]3s^{-1/3}ds=v\cdot dv[/tex]

[tex]\Rightarrow 3*\frac{3}{2}s^{2/3}+C=v+C[/tex]

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

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

Thanks for the help so far blochwave!
 
  • #4
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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:
  • #5
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It says "car starts from rest" which implies that at s(0)=0.
 
  • #6
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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
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My instructor e-mailed me a long hint. I can't seem to make sense of it. It says :

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

Then [itex]a=\frac{dv}{ds}*v[/itex] (1)

Integrate (1) to get [itex]v=3s^{1/3}[/itex]

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

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

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

Not [itex]v=3s^{1/3}[/itex]


arrgh!
 
  • #8
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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?:redface:
 
  • #9
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Can anybody help me out here? I really need to get this last one. Thanks!!
 
  • #10
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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
 

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