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Finding acceleration when velocity is given as function of displacement

  • Thread starter maunder
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  • #1
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Homework Statement

The question is:
The velocity of a particle is given as a function of its position s:

v = (A - 4 s2)1/2

where s is displacement (in m), and v is velocity (in m/s). A is a constant but you do not need to know its value to answer this question.

What is the acceleration (in m/s2) when s is as given below?

S=1.9m

The attempt at a solution
i have tried integrating using the fact that v=ds/dt
this gets 1/((A - 4 s2)1/2)ds=dt.

I integrated first by using partial fractions then using the fact that 1/(A2-x2)1/2=arcsin(x/a)
However whatever i do i end up with the constant A still in there??
 

Answers and Replies

  • #2
diazona
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That formula can't be right - the units don't work out. If s has units of length, [tex](A - 4s^2)^{1/2}[/tex] also has units of length, not velocity. So something must be missing.

Anyway, what do you get when you integrate velocity?
[tex]\int v(t) \mathrm{d}t = ?[/tex]
What about when you differentiate velocity?
 
  • #3
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the integral i get is [tex]\frac{-ln(2s-\sqrt{A})}{4\sqrt{A}}+\frac{ln(2s+\sqrt{A})}{4\sqrt{A}}[/tex] by using integration by partial fractions or
[tex]\frac{1}{2}tan^{-1}\frac{2s}{\sqrt{a-4s^{2}}}[/tex]

The function isn't wrong, the velocity is given in terms of its position.
I'm thinking what will happen is the A will cancel out when i do a definite integral, but i cant ever get A by itself, it is always a co-efficient of s.

when you say differentiate velocity, i'm assuming you mean with respect to time.
When you differentiate the RHS is dissappears because it is independent of t in the first place.
 
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  • #4
gabbagabbahey
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When you differentiate the RHS is dissappears because it is independent of t in the first place.
Careful, just because there is no explicit time dependence, does not mean that there is no implicit time dependence. The displacement 's' will depend on time won't it?....You will need to use the chain rule on this one.
 
  • #5
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i dont understand?? how would i express s in terms of t
 
  • #6
gabbagabbahey
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i dont understand?? how would i express s in terms of t
You don't need to; just use the chain rule:

[tex]\frac{d}{dt}v(s)=\frac{dv}{ds}\frac{ds}{dt}[/tex]

you can easily calculate dv/ds, and you should already know what ds/dt is :wink:
 
  • #7
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dont worry, i got it.
[tex]a=\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}=v\frac{dv}{ds}[/tex]
[tex]\frac{dv}{ds}=\frac{-4s}{\sqrt{A-4s^{2}}}[/tex]
[tex]v\frac{dv}{ds}=-4s=a[/tex]

I'd tried this before, but i must of stuffed it up. Thanks guys. I cant beleave how easy it was.
 
  • #8
Redbelly98
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That formula can't be right - the units don't work out. If s has units of length, [tex](A - 4s^2)^{1/2}[/tex] also has units of length, not velocity. So something must be missing.
Since the OP explicitly says s is the displacement in m, we can think of it as

s = displacement / 1m​

and similarly for v. So all quantities are dimensionless, in that sense.
 
  • #9
diazona
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Since the OP explicitly says s is the displacement in m, we can think of it as

s = displacement / 1m​

and similarly for v. So all quantities are dimensionless, in that sense.
True, I missed that... but then the OP says "S = 1.9m" (which I assumed should have been lowercase s), so there's still an inconsistency.
 
  • #10
Redbelly98
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True, I missed that... but then the OP says "S = 1.9m" (which I assumed should have been lowercase s), so there's still an inconsistency.
You're right, that would make the displacement equal to 1.9m2. Gaaaah! :eek:
 
  • #11
diazona
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You're right, that would make the displacement equal to 1.9m2. Gaaaah! :eek:
lol, I know the feeling :biggrin: It always irritates me when people don't keep track of the units properly.
 

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