# Having difficulty with concept of turning points

1. Feb 13, 2015

### Marcus27

1. The problem statement, all variables and given/known data

I understand how to solve for stationary points, and then take the second derivative and input the values of x to determine the nature of the stationary points, if x > 0 then the stationary point is a minimum and if x < 0 then the stationary point is a maximum. What I am having difficulty with is the concept.
2. Relevant equations

If f(x) = x^3 -3x^2 + 4
f'(x) = 3x^2 -6x
f''(x) = 6x - 6
Lets say that f(x) gives the distance travelled by a car after x seconds, so f'(x) would give a speed graph, and f''(x) would be an acceleration graph. However, solving for the stationary point of the graph f(x)
f'(x) = 0
3x^2 -6x = 0
3x(x-2) = 0
Stationary points are (0,4) and (2,0)
Plugging these values into the acceleration function,
f''(0)= 6(0) - 6 = -6
f''(2) = 6(2) -6 = 6

This is all well and good but I do not understand how a car can be travelling at 0 speed and have a negative or positive acceleration.
3. The attempt at a solution

2. Feb 13, 2015

### Dick

Acceleration is the rate of change of speed. I don't see any problem with having a speed of 0 and a rate of change of speed that's not 0.

3. Feb 13, 2015

### Marcus27

Newton's laws state an object will remain at rest or in motion until an unbalanced force acts upon it, so does the acceleration not effect the motion of the car and therefore its speed?.

4. Feb 13, 2015

### Dick

Then having a nonzero acceleration means a unbalanced force is acting on the car. What's the problem? Of course having a nonzero acceleration affects the cars motion and speed.

5. Feb 13, 2015

### Marcus27

Oh, I think I understand now the derivative of distance as a function of time = instantaneous speed, for some reason I was treating it as constant speed. So, if a car starts of at -10m/s and accelerates at 5m/s^2 then at time = 2 seconds it's speed is equal to zero while it maintains an acceleration of 5 m/s^2. Thank you for helping me clarify this. :)

6. Feb 13, 2015

### Dick

Exactly. Wish I could have explained it that well.