Acceleration when velocity is a function of distance

1. Apr 29, 2008

Fyei

1. The problem statement, all variables and given/known data
A car has a variable veloctiy given as a funtion of distance s by 1 +2s. What its accleration when it has travelled 10m and how far has it travelled in 4s.

2. Relevant equations

$$a(s)=v\frac{dv}{ds}$$(i think?)

$$a=\frac{d^2s}{dt^2}$$

3. The attempt at a solution

accleration at 10m

$$a(s)=v\frac{dv}{ds}=(2s+1)(2) = (4s+2)$$

$$a(10) = 42ms^-2$$

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: Apr 29, 2008
2. Apr 29, 2008

tiny-tim

Welcome to PF!

Hi Fyei! Welcome to PF!

Nice solution for the first part!

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

You'll have to solve a differential equation in t.

3. Apr 29, 2008

Fyei

yes s means seconds, sorry about the confusion

and differential in t, you mean something along the lines

$$\frac{dv}{dt}=\frac{d^2s}{dt^2}$$?

4. Apr 29, 2008

tiny-tim

Hi Fyei!

No … more like ds/dt = 1 + 2s.

5. Apr 29, 2008

Fyei

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 = $$e^{\int-2ds} = e^{-2s}$$

$$s=\frac{\int1*e^{-2s}ds}{e^{-2s}}$$

therefore

$$s=\frac{1*-2e^{-2s}}{e^{-2s}}$$

$$s=-2$$

which is obviously wrong

Last edited: Apr 29, 2008
6. Apr 29, 2008

DavidWhitbeck

Separate and integrate if you want s(t)--

$$\int ds/(1+2s)=\int dt$$

$$s=\frac{1}{2}(1-e^{2t})$$

7. Apr 29, 2008

kamerling

Separation of variables is easier. your method should work but in this equation:

$$s=\frac{1*-2e^{-2s}}{e^{-2s}}$$

- you should have had t instead of s as the dependent variable
- you forgot about the integration constant
- the integral of $e^{-2t}$ is $-\frac{1}{2} e^{-2t}$

8. Apr 29, 2008

Fyei

thank you, I'm feeling a bit silly at all my school boy errors