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 [-1,5].

Question

So I found the antiderivative of the function, which gave me t^3/3-5^2/2+4x+c. Then plugging in the given points and using the Fundamental Thm of Calculus, I got 6.

I know this is either the displacement or the distance traveled, but I'm not sure which. Can you help me understand which it is, and how to find the other value as well?

Answer 1

let x be distance
then dx/dt is velocity
so INT dx/dt dt has to be displacement

your antiderivative should be t^3/3-5t^2/2+4t+c.

The area under the curve of v(t) over time is displacement. If at any point that area is below the displacement =0 axis, then that "negative" area subtracts. For distance, you would want to add.
so you can integrate the abs value function of velocity to get distance. Or you can notice for the time -1 to zero, the function will be negative, figure that and add it to the other integral to get DISTANCE.

I did not check your displacement function to see if you got the right answer. Recheck it.