# Non constant acceleration

1. Apr 25, 2012

### albert12345

1. The problem statement, all variables and given/known data
A car accelerates with the acceleration a(s)=k*s^n m/s^2
s= distance in m
k = 5,8
n=0,7351
The car starts the acceleration from 0 m/s

How far has the car travelled when it reaches the speed 4,61m/s?

3. The attempt at a solution
When i anti derivate one time i will get dv/dt. Do i put in the distance after that?

2. Apr 25, 2012

### albert12345

Where x is the distance "s"
so I get this:

(1/2)*(x^2) = v2^2 - V1^2

Where v2 is the final speed and v1 is the starting speed=0..

Am i right or wrong?

3. Apr 25, 2012

### ehild

What do you mean? dv/dt =a, the acceleration.

You have a differential equation dv(s(t))/dt=k*s^n. Write up the left-hand side with applying the chain rule:

dv/dt=dv/ds ds/dt. ds/dt=v, so v dv/ds =ks^n. This is a differential equation for v(s), easy to solve.

ehild

4. Apr 25, 2012

### albert12345

So I just plug in my values here, or do i have to antiderivate first?

v dv/ds =ks^n

5. Apr 25, 2012

### albert12345

So I just plug in my values here and solve for s, or do i have to antiderivate first?

v dv/ds =ks^n

6. Apr 25, 2012

### ehild

How do you want to solve for s? Show your attempt.

ehild