Question

A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s.

Answer 1

Answer:

0 m/s , 3 m/s , 2 m/s^2

Explanation:

Given : s(t) = ( t^2 - 6t + 5)

v(t) = ds / dt = 2t - 6

s(0) = 5 m

s(6) = (6)^2 - 6*6 + 5 = 5 m

Vavg = ( s(6) - s(0) ) / 2 = 0 m\s

Find the turning point of particle:

ds/dt = 0 = 2t - 6

t = 3 sec

s(3) = 3^2 -6*3 + 5 = - 4

Total distance = 5 - (-4) + (5 - (-4)) = 18 m

Total time = 6s

Average speed = Total distance / Total time = 18 / 6 = 3 m/s

Taking derivative of v(t) to obtain a(t)

a (t) = dv(t) / dt = 2 m/s^2