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

Question

. a) What is the average velocity of the particle over the interval 0 ≤ t ≤ ? 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

a) v_av = (s(5) - s(0))/(5 - 0) = (- 10 - 0)/5 = - 2

b) v'(t) = 3 + 8t - 3t²; v(2) = 3 + 16 - 12 = 7 > 0. So, at t = 2 particle moves to the right.

c) Particle change direction when v(t) = 0; 3 + 8t - 3t² = 0; 3t² - 8t - 3 = 0; (3t + 1)(t - 3) = 0; t = 0.

So, at t = 3 direction changed

d) At t = 4 v(4) = 3 + 32 - 48 = - 13 < 0 and accelaration a(t) = v'(t) = 8 - 6t. So, a(4) = - 16 < 0.

Because, speed is increasing at t = 4