Question

A particle moves along a straight line. For 0 ≤ t ≤ 5, the position of the particle is given by s(t)=3t+4t^2-t^3.

a) What is the average velocity of the particle over the interval 0 ≤ t ≤ 4?
b) Find a general expression for the velocity of the particle as a function of time, v(t), and determine the value of v at t = 2. Is the particle traveling to the left or the right at this point? Justify your answer.
c) Find all times t during the interval 0 ≤ t ≤ 5 that the particle changes direction. Justify your answer.
d) At t = 4, is the speed of the particle increasing or decreasing? Justify your answer.

Answer 1

The average velocity over 0 ≤ t ≤ 4 is 2 m/s. The particle's velocity function is v(t) = 3 + 8t - 3t^2, and at t = 2, v is 7 m/s meaning the particle is moving to the right. The particle changes direction at t = 0, t = 2, and t = 4, and at t = 4, the speed is decreasing due to negative acceleration.

Step-by-step explanation:

To answer a student's question about the motion of a particle along a straight line with the position function s(t) = 3t + 4t^2 - t^3 for 0 ≤ t ≤ 5:

1. Average velocity over the interval 0 ≤ t ≤ 4:
   To find the average velocity, we use the formula:
   average velocity = (s(4) - s(0)) / (4 - 0)
   Plugging in the position function:
   average velocity = (3(4) + 4(4)^2 - (4)^3 - (3(0) + 4(0)^2 - (0)^3)) / 4
   Calculating this gives an average velocity of 2 m/s.
2. Velocity as a function of time, v(t):
   The velocity function is the derivative of the position function:
   v(t) = s'(t) = 3 + 8t - 3t^2
   The value of v at t = 2 is:
   v(2) = 3 + 8(2) - 3(2)^2
   v(2) = 3 + 16 - 12
   v(2) = 7 m/s
   The positive value indicates the particle is traveling to the right.
3. Times when the particle changes direction:
   To find when the particle changes direction, we set the velocity function equal to zero and solve for t in the interval 0 ≤ t ≤ 5. The particle changes direction when the velocity changes sign, which occurs at t = 0, t = 2, and t = 4.
4. At t = 4, we check if the speed of the particle is increasing or decreasing by finding the acceleration and assessing its sign:
   a(t) = v'(t) = 8 - 6t
   a(4) = 8 - 6(4)
   a(4) = 8 - 24
   a(4) = -16 m/s^2
   The negative acceleration means the speed is decreasing.