Drew asked in Science & MathematicsPhysics · 2 months ago

# 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
• 2 months ago

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 7
2 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