Motion problems: finding the maximum acceleration | AP Calculus AB | Khan Academy

A particle moves along the x-axis so that at any time T greater than or equal to zero, its velocity is given by ( V(T) = T^3 + 6T^2 + 2T ).

At what value of T does the particle obtain its maximum acceleration? So we want to figure out when it obtains its maximum acceleration.

Let’s just review what they gave us. They gave us velocity as a function of time. So let’s just remind ourselves: if we have, let’s say, our position is a function of time, so let’s say ( X(T) ) is position as a function of time, then if we were to take the derivative of that, ( X'(T) ), well, that’s going to be the rate of change of position with respect to time, or the velocity as a function of time.

If we were to take the derivative of our velocity, then that’s going to be the rate of change of velocity with respect to time—well, that’s going to be acceleration as a function of time. So they give us velocity. From velocity, we can figure out acceleration.

Let me just rewrite that. So we know that ( V(T) = T^3 + 6T^2 + 2T ). From that, we can figure out the acceleration as a function of time, which is just going to be the derivative with respect to T of the velocity.

So just use the power rule a bunch. That’s going to be this is a third power right there: ( 3T^2 + 12T + 2 ). So that’s our acceleration as a function of time. We want to figure out when we obtain our maximum acceleration.

Just inspecting this acceleration function here, we see it's quadratic; it has a second-degree polynomial. We have a negative coefficient out in front of the highest degree term, in front of the quadratic second-degree term, so it is going to be a downward opening parabola.

Let me draw in the same color. So it is going to have that general shape, and it will indeed take on a maximum value. But how do we figure out that maximum value? Well, that maximum value is going to happen when the acceleration value, when the slope of its tangent line is equal to zero.

We could also verify that it is concave downwards at that point using the second derivative test by showing that the second derivative is negative there. So let’s do that; let’s look at the first and second derivatives of our acceleration function.

I’ll switch colors; that one’s actually a little bit hard to see. The first derivative, the rate of change of acceleration, is going to be equal to: so this is ( -6T + 12 ). Now let’s think about when this thing equals zero. Well, if we subtract 12 from both sides, we get ( -6T = -12 ).

Divide both sides by -6; you get ( T = 2 ). So a couple of things: you could just say, “All right, look, I know that this is a downward opening parabola right over here. I have a negative coefficient on my second-degree term. I know that the slope of the tangent line here is zero at ( T = 2 ), so that’s going to be my maximum point.”

Or you could go a little bit further; you can take the second derivative. Let’s do that just for kicks. So we could take the second derivative of our acceleration function. This is going to be equal to 6, right? The derivative of ( -6T ) is 6, and the derivative of a constant is just zero.

So this thing, the second derivative, is always negative. So we are always concave downward. And so by the second derivative test at ( T = 2 ), well, at ( T = 2 ), our second derivative of our acceleration function is going to be negative.

And so we know that this is our maximum value, or max, at ( T = 2 ). So at what value of T does the particle obtain its maximum acceleration? At ( T = 2 ).