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# The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = 9t − 5, 0 ≤ t ≤ 3?

The velocity function, in feet per second, is given for a particle moving along a straight line.

v(t) = 9t − 5, 0 ≤ t ≤ 3

(a) Find the displacement= 25.5 ft

I understand how to find displacement.

(b) Find the total distance that the particle travels over the given interval.

The total distance is giving me trouble.

### 2 Answers

- MathguyLv 52 months agoFavorite Answer
First up... don't forget to choose a Best Answer...

Total distance is just the ∫ | v(t) | dt

That is , the integral of the absolute value of the velocity, from 0 to 3 I get an answer of a bit over 28 ft. from a Graphing Calculator...

Or you can split the | v | into pieces... it is = 0 at t = 5/9 sec so 2 integrals ...

-1 * ∫ ( 9t- 5 )dt from 0 to 5/9 PLUS ∫ ( 9t - 5 ) dt from 5/9 to 3 if you need to show all the steps....

- ?Lv 72 months ago
distance traveled s = int v dt= 9/2 t^2-5t. Put the limits

s= 9/2 9-15= 81/2-15= (81-30)/2= 51/2=25.5 ft