Acceleration when velocity is a function of distance

  • Thread starter Fyei
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  • #1
Fyei
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Homework Statement


A car has a variable veloctiy given as a funtion of distance s by 1 +2s. What its accleration when it has traveled 10m and how far has it traveled in 4s.

Homework Equations



[tex]a(s)=v\frac{dv}{ds}[/tex](i think?)

[tex]a=\frac{d^2s}{dt^2}[/tex]


The Attempt at a Solution




accleration at 10m

[tex]a(s)=v\frac{dv}{ds}=(2s+1)(2) = (4s+2)[/tex]

[tex] a(10) = 42ms^-2[/tex]

however i am unsure how to do the second part as velocity is already a function of distance, is it solved using the chain rule?
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Homework Statement


A car has a variable veloctiy given as a funtion of distance s by 1 +2s. What its accleration when it has traveled 10m and how far has it traveled in 4s.

Hi Fyei! Welcome to PF! :smile:

Nice solution for the first part! :biggrin:

In the second part, the 4s means 4 seconds, doesn't it?

You'll have to solve a differential equation in t. :smile:
 
  • #3
Fyei
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yes s means seconds, sorry about the confusion

and differential in t, you mean something along the lines

[tex]\frac{dv}{dt}=\frac{d^2s}{dt^2}[/tex]?
 
  • #4
tiny-tim
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Hi Fyei! :smile:

No … more like ds/dt = 1 + 2s. :smile:
 
  • #5
Fyei
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I tried to do that but I must be solving it wrong

ds/dt = 1 + 2s

therefore

ds/dt-2s = 1

now in the form dy/dx+Py=Q

therefore can be solved with intergrating factors

thefore If = [tex]e^{\int-2ds} = e^{-2s}[/tex]

[tex]s=\frac{\int1*e^{-2s}ds}{e^{-2s}}[/tex]

therefore

[tex]s=\frac{1*-2e^{-2s}}{e^{-2s}}[/tex]

[tex]s=-2[/tex]

which is obviously wrong
 
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  • #6
DavidWhitbeck
351
1
Separate and integrate if you want s(t)--

[tex]\int ds/(1+2s)=\int dt[/tex]

[tex]s=\frac{1}{2}(1-e^{2t})[/tex]
 
  • #7
kamerling
454
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Separation of variables is easier. your method should work but in this equation:

[tex]
s=\frac{1*-2e^{-2s}}{e^{-2s}}
[/tex]

- you should have had t instead of s as the dependent variable
- you forgot about the integration constant
- the integral of [itex] e^{-2t} [/itex] is [itex] -\frac{1}{2} e^{-2t} [/itex]
 
  • #8
Fyei
5
0
thank you, I'm feeling a bit silly at all my school boy errors
 

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