The velocity function is for a particle moving along a line. Find the displacement and the distance traveled?

The velocity function is v(t) = - t^2 + 6t - 8 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-3,5].

displacement = -66.67

distance traveled = ?

I found the displacement, but finding the distance traveled is going me trouble. A solution would be appreciated. Thanks


To find the total distance traveled find the definite integral of the absolute value of the velocity.

Graphing y = -x^2 + 6x + 8, we see that the velocity is positive if 2 <?t < 4 and negative if 0 < 2 < t or t > 4.

So on the interval [-3, 5], the velocity is negative between t = -3 and t = 2, positive between t = 2 and t = 4, and negative again between t = 4 and t = 5.

Instead of integrating the absolute value of the velocity over the interval, we can do the following:

-∫ {-3, 2} v(t) dt + ∫?{2, 4} v(t) dt - ∫ {4, 5} v(t) dt = 69.333....

The total distance traveled is 69 1/3.


First understand the difference. When I go to school, have class, and go back home, my displacement from my origin is zero as I have no net displacement. But I did have to travel 10 miles to get there, 10 miles to get back.

Now assume my velocity is positive going to school, what is it going back? it's negative since it gives opposite direction.

So if I took the absolute value of the displacement, and added them all up, I will get distance traveled.

Taking the equation, we see when I put in -3, I get -35 in return. When will it be zero? Factor and see...

-(x-2)(x-4)=0... x is 2 and 4.

So between -3 and 2, it's negative, between 2 and 4 it's positive, and from 4 to 5 it's negative. Integrate using these values, but always disregard the sign as we are only looking at the absolute values

You should get something like 69.33... Don't have my calc on me so did it on a graphing program