Particle Motion Problem

Question

The position of a particle is given by the function: S(t)=t^3-6t^2+9t where t is in seconds, S(t) is in meters and 0 less than or equal to t less than or equal to 5. Find the times when particle is at rest. Then find the position of the particle when it is at rest and at t=0 and t=5. Draw a diagram to represent the motion of the particle. Label times and positions of the particle and the direction in which it is traveling.

Answer 1

If s(t) = t³ - 6t² + 9t then, since v(t) = s'(t), v(t) = 3t² - 12t + 9

The particle has a velocity of zero when it is "at rest" so, set v(t) equal to zero:

0 = 3t² - 12t + 9

0 = t² - 4t + 3

0 = (t - 1)(t - 3)

t = 1 and t = 3

So the particle is at rest at times of 1 second and 3 seconds

Its position at time t = 1 is s(1)

s(1) = (1)³ - 6(1)² + 9(1) = 1 - 6 + 9 = 4 so the particle is 4 meters to the right of the starting position (since at t = 0, s = 0)

Its position at time t = 3 is s(3)

s(3) = (3)³ - 6(3)² + 9(3) = 27 - 54 + 27 = 0 so the particle is back at the starting position

Its position at time t = 0 is s(0)

s(0) = (0)³ - 6(0)² + 9(0) = 0

Its position at time t = 5 is s(5)

s(5) = (5)³ - 6(5)² + 9(5) = 125 - 150 + 45 = 20 so the particle is 20 meters to the right of the starting position

A diagram might look like this: