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 Question from Mary, a student: Velocity of a function (which is the first derivative of its position) is defined over the interval 0 to 12 using the following piecewise function: v(t)=-1 from 0 to 4, v(t)=x-5 from (4 to 8 and v(t)=-x+11 from (8 to 12. At what value of t is the maximum acceleration?

Hi Mary.

The acceleration is the derivative of a velocity curve. But derivatives are only defined at points where the limit of the function exists and is equal on both sides of the point. What this means is that if you graph a function and there are "kinks" or corners, then no slope or derivative is defined for those particular points.

Let's examine the parts of your piecewise function:

Time t = 0 is an end point, so there is no defined limit on the left side, thus there is no acceleration defined.
Time 0 < t < 4 has a constant value, so the slope is 0, thus the acceration is 0.
At time t = 4, the left side limit is 0 and the right side is 1, so they aren't equal and so no acceleration is defined.
Time 5 < t < 8 has a constant slope of 1, thus the acceleration here is 1.
At time t = 8, the left side limit is 1 and the right side is -1, so they aren't equal and thus no acceleration is defined.
Time 8 < t < 12 has a constant slope of -1, thus the acceleration here is -1.
Time t = 12 is an end point, hence there is no acceleration defined.

So from a mathematical perspective, the maximum known acceleration is 1, in the open interval (5, 8).

However, if the question is more of a physics problem, then the "points" t = 4, t = 8 may be intended to mean "small time periods" rather than points in time (!) If so, the question is posed rather poorly, but you could think of it this way:

At time t = 4, the speed changes from 0 to 1 in a tiny amount of time. At time t = 8 the speed changes from 1 to -1 in the same amount of time. So in the first case it changes by 1 unit and in the second by -2 units. If the person who gave you this question is looking for a "physics" kind of answer, then he or she is probably hoping you'll say time t = 8 seconds.

But you know better: the way the question is worded, there is no mathematically-defined acceleration at that time!

Cheers,
Stephen La Rocque

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.