If the position of the particle is : 1/3x^3 + 2x^2 - 5x + 1. What is the total distance traveled by the particle between 0 and 2 seconds.

Answer 1

The derivative is x²+4x-5, which factors into (x-1)(x+5). This means the only critical point in this interval is x=1. Using the test points x=0 and x=2, we can see that this function is decreasing (moving backwards) on [0,1) and increasing (moving forward) on (1,2]. Since distance is not dependent on direction, we want to integrate the absolute value of the velocity, which is the same as integrating -(x²+4x-5) from 0 to 1 and x²+4x-5 from 1 to 2 and adding them together. A quick computation yields that it's traveled 6 units.