Aparticle moves along a straight line with a velocity in millimeters per second given by v = 301 - 16t2 where tis in seconds. calculate the net displacement δs and total distance d traveled during the first 5 seconds of motion.
answers:
δs = m
d = m

a) Δs= 1255 m

b) D = 1255 m

Explanation:

Velocity, v(t) = 301 - 6t²

a) Displacement is given by integral of v(t) from 0 to 5.

b) v(t) = 301 - 6t²

  301 - 6t² = 0

  t = 7.08 s

  So 301 - 6t² is negative when t is greater than 7.08 s.

  So from 0 to 5 seconds displacement is equal to distance.

   Distance is given by integral of v(t) from 0 to 5.