Question

A particle 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

Answer 1

Answer:

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.

   $\Delta s=\int_{0}^{5}\left ( 301-6t^2\right )=\left [ 301t-2t^3\right ]_0^5=301\times 5-2\times 5^3=1255m$

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.

   $D=\int_{0}^{5}\left ( 301-6t^2\right )=\left [ 301t-2t^3\right ]_0^5=301\times 5-2\times 5^3=1255m$