Having difficulty with concept of turning points

[Marcus27]

Homework Statement

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.

Homework 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 traveled 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 traveling at 0 speed and have a negative or positive acceleration.

The Attempt at a Solution

 

[Dick]

Homework Statement

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.

Homework 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 traveled 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 traveling at 0 speed and have a negative or positive acceleration.

The Attempt at a Solution

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.

 

[Marcus27]

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.

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?.

 

[Dick]

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?.

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.

 

[Marcus27]

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.

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. :)

 

[Dick]

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. :)

Exactly. Wish I could have explained it that well.