# 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.

Relevance

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....

• 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