# Acceleration when velocity is a function of distance

## Homework Statement

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.

## Homework Equations

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

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

## 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?

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tiny-tim
Homework Helper
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 travelled 10m and how far has it travelled in 4s.
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.

yes s means seconds, sorry about the confusion

and differential in t, you mean something along the lines

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

tiny-tim
Homework Helper
Hi Fyei!

No … more like ds/dt = 1 + 2s.

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

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

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

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

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 $-\frac{1}{2} e^{-2t}$

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