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# Motion problems: finding the maximum acceleration

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.B (LO)
,
FUN‑4.B.1 (EK)
,
FUN‑4.C (LO)
,
FUN‑4.C.1 (EK)

## Video transcript

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 of T is equal to negative T to the third power plus TC plus 6t squared plus 2t and what value of T does the particle obtain its maximum acceleration so we want to figure out what does it obtain its maximum acceleration so let's just review with the games they gave us velocity is 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 of T is position as a function of time then if we were to take the derivative of that so X prime of 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 and 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 so from velocity we can figure out acceleration so let me just rewrite that so we know that V of T is equal to negative T to the third power plus 6t squared plus 2t and so 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 so that's going to be this is a third power right there so negative 3 T squared plus 2 times 6 is 12 T to the first plus 2 so that's our acceleration as a function of time and we want to figure out when we obtain our maximum acceleration and just inspecting this acceleration function here we see it's a quadratic it has the second degree polynomial we have a negative coefficient out in front of the highest degree term in front of the quad the second degree term so there is going to be a downward-opening parabola so it is going to be a downward-opening let me draw it in the same color so it is going to have that general shape and so it will indeed take on 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 values when the when it's when the slope of its tangent line is equal to when the slope of its tangent line is equal to zero and 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 of our acceleration function so and I'll switch colors that one's actually a little bit hard to see so the first derivative the rate of change of acceleration is going to be equal to so this is negative 6t plus twelve now let's think about when does this thing equal zero well if we subtract 12 from both sides we get negative 6t is equal to negative 12 divide both sides by negative six you get T is equal to two so a couple of things you could just say alright look I know that this is a downward-opening Quadra a parabola right over here have a negative coefficient on my second degree term I know that the slope of the tangent line here is zero at T equals two so that's going to be my maximum point or you can go a little bit further you can take the second derivative U let's do that just for kicks so we could take the second derivative of our acceleration function so this is going to be equal to negative six right the derivative of negative 6t is negative six the derivative of constant is just zero so this thing the second derivative is always negative so we are always always concave concave downward downward and so by the second derivative test at T equals two well at T equals two our second derivative our acceleration function is going to be negative and so we know that this is our maximum value or max at T is equal to two so at what value of T does the particle obtain its maximum acceleration at T is equal to at T is equal to two