Acceleration when velocity is a function of distance

In summary, the car has an acceleration of 42 ms^-2 when it has traveled 10m and has traveled a distance of -2m in 4s. The second part can be solved using separation of variables, yielding s(t) = 1/2(1-e^2t).
  • #1
Fyei
5
0

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:
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  • #2
Welcome to PF!

Fyei said:

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
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
Hi Fyei! :smile:

No … more like ds/dt = 1 + 2s. :smile:
 
  • #5
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
 
Last edited:
  • #6
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
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
thank you, I'm feeling a bit silly at all my school boy errors
 

What is acceleration when velocity is a function of distance?

Acceleration when velocity is a function of distance refers to the rate of change of velocity with respect to distance. In other words, it is the change in velocity over a certain distance.

How is acceleration calculated when velocity is a function of distance?

Acceleration can be calculated by taking the derivative of the velocity function with respect to distance. This will give the instantaneous acceleration at a specific point along the distance.

What is the difference between acceleration and velocity when velocity is a function of distance?

Velocity is a vector quantity that measures the rate of change of an object's position over time, while acceleration is a vector quantity that measures the rate of change of an object's velocity over time. When velocity is a function of distance, acceleration is determined by the change in velocity over a specific distance, rather than time.

What is the relationship between acceleration and distance when velocity is a function of distance?

The relationship between acceleration and distance is directly proportional when velocity is a function of distance. This means that as the distance increases, the acceleration also increases, and vice versa.

Can acceleration be negative when velocity is a function of distance?

Yes, acceleration can be negative when velocity is a function of distance. This indicates that the object is slowing down as it moves further along the distance, or that it is moving in the opposite direction of the positive direction defined by the distance.

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