This is the distance that the particle traveled.
Example 5: Given the position function x(t) = t^4 − 8t^2 , find the distance that the particle travels from t = 0
to t = 4.
First, find the first derivative (v(t) = 4t^3 − 16t) and set it equal to zero.
4 t^3 − 16t = 0 4t(t^2 − 4) = 0 t = 0, 2, −2So we need to divide the time interval into t = 0 to t = 2 and t = 2 to t = 4.
|x(2) − x(0)| + |x(4) − x(2) = 16 + 144 = 160Here are some solved problems. Do each problem, covering the answer first, then checking your answer.
PROBLEM 1. Find the velocity and acceleration of a particle whose position function is x(t) = 2t^3 − 21t^2 +
60 t + 3, for t > 0.
Answer: Find the first two derivatives.
v(t) = 6t^2 − 42t + 60a(t) = 12t − 42PROBLEM 2. Given the position function in problem 1, find when the particle’s speed is increasing.
Answer: First, set v(t) = 0.
6 t^2 − 42t + 60 = 0t^2 − 7t + 10 = 0(t − 2)(t − 5) = 0t = 2, t = 5You should be able to determine that the velocity is positive from 0 < t < 2, negative from 2 < t < 5, and
positive again from t > 5.
Now, set a(t) = 0.
12 t − 42 = 0t = You should be able to determine that the acceleration is negative from 0 < t < and positive from t > .