a(t) = v′(t) = 6t − 24So the acceleration will be negative when t < 4, and positive when t > 4.
So we have
Time Velocity Acceleration0 < t < 2 Positive Negative2 < t < 4 Negative Negative4 < t < 6 Negative Positivet > 6 Positive PositiveWhenever the velocity and acceleration have opposite signs, the particle is slowing down. Here the
particle is slowing down during the first two seconds (0 < t < 2) and between the fourth and sixth seconds
(4 < t < 6).
Another typical question you’ll be asked is to find the distance a particle has traveled from one time to
another. This is the distance that the particle has covered without regard to the sign, not just the
displacement. In other words, if the particle had an odometer on it, what would it read? Usually, all you
have to do is plug the two times into the position function and find the difference.
Example 4: How far does a particle travel between the eighth and tenth seconds if its position function is
x(t) = t^2 − 6t?
Find x(10) − x(8) = (100 − 60) − (64 − 48) = 24.
Be careful about one very important thing: If the velocity changes sign during the problem’s time
interval, you’ll get the wrong answer if you simply follow the method in the paragraph above. For
example, suppose we had the same position function as above but we wanted to find the distance that the
particle travels from t = 2 to t = 4.
x(4) − x(2) = (−8) − (−8) = 0This is wrong. The particle travels from −8 back to −8, but it hasn’t stood still. To fix this problem,
divide the time interval into the time when the velocity is negative and the time when the velocity is
positive, and add the absolute values of each distance. Here the velocity is v(t) = 2t − 6. The velocity is
negative when t < 3 and positive when t > 3. So we find the absolute value of the distance traveled from t
= 2 to t = 3, and add to that the absolute value of the distance traveled from t = 3 to t = 4.
Because x(t) = t^2 − 6t,
|x(3) − x(2)| + |x(4) − x(3)| = |−9 + 8| + |−8 + 9| = 2