Integral Help! Calculus

Question

The velocity function is v(t)=-t^2+5t-4 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-3,5].

Answer 1

1) For the displacement, you simply integrate the velocity function choosing the lower limit as -3 and upper limit as 5. This gives you the net change in your position.

2) For the total distance traveled, you need to be careful. It would be a very good idea to plot the velocity vs time graph, and determine the regions where the velocity function is negative and positive. You will see that from -∞ to +1, and from +4 to +∞, the velocity function is negative. In order to find the total distance traveled you need to find the integral of the absolute value of the velocity function. So, for your problem, you will need to calculate first the integral of (t^2-5t+4) from -3 to 1, then the integral of (-t^2+5t-4) from 1 to 4, and finally the integral of (t^2-5t+4) from 4 to 5 (Hence the sign changes in the functions. In the regions from -3 to 1 and from 4 to 5, the absolute value of the velocity function is (t^2-5t+4)). Then, the sum of these three integrals will give you the total distance traveled. Please let me know if you have any trouble taking the integral. As a side note, displacement is a vector and distance traveled is a scalar. In cases where the velocity changes sign and/or changes direction, the total distance traveled is bigger than the magnitude of the displacement. For example, in a circular motion, after the completion of one revolution the displacement becomes zero (you end up being at your initial position), whereas the distance traveled is 2ΠR (circumference of the circle)

Answer 2