For 0 ≤ 𝑡 ≤ 12, a particle moves along the x-axis. The velocity of the particle at time t is given by v(t)=t^2 −10t+21

Question

For 0 ≤ 𝑡 ≤ 12, a particle moves along the x-axis. The velocity of the particle at time t is given by v(t)=t^2 −10t+21

a. For 0 ≤ 𝑡 ≤ 12, when is the particle moving to the left?

b List all values of t on the interval 0 ≤ 𝑡 ≤ 12 when the particle changes directions.

c Find the acceleration of the particle at any time t.

d When t = 4, determine if the speed of the particle is increasing, decreasing or neither. Explain your reasoning.

Answer 1

Hello, Sara. Let's do this one!

v(t) = t² - 10t + 21 for 0 ≤ t ≤ 12

(a) Particle moves left when v(t) < 0 => t² - 10t + 21 < 0

(t - 7)(t - 3) < 0

=> t - 7 < 0 and t - 3 > 0 OR t - 7 > 0 and t - 3 < 0

=> t < 7 and t > 3 OR t > 7 and t < 3

=> 3 < t < 7 (the second alternative is impossible in real numbers)

(b) Since v(t) is a quadratic, there is only value of t where it changes direction, namely, at minimum velocity.

The minimum occurs on the axis of symmetry: x =-b/2a = -(-10) / (2 x 1) = +5 seconds

(c) By definition, acceleration, a(t) = dv/dt = 2t -10

(d) From part (c): a(4) = 2 x 4 - 10 = -2

Since acceleration is negative, the particle must be slowing down, i.e., its velocity is decreasing.