Solve the following question.

Question

The position of a particle moving along the 𝑥-axis is given by 𝑠(𝑡) = 5 + 3𝑡^2 − 2𝑡^3 meters for 𝑡 ≥ 0 seconds.

A) Find the initial position, 𝑡 = 0, of the particle.

B) Find the displacement for 1 ≤ 𝑡 ≤ 3.

C) Find the average velocity for 1 ≤ 𝑡 ≤ 3.

D) Find the velocity at 𝑡 =2 seconds. Find the speed of the particle at 𝑡 = 2 seconds. Is the ball moving forward or backward at 𝑡 = 2 seconds? Justify.

E) How long does it take for the velocity to reach −36 m/s?

F) Find the acceleration at 𝑡 = 2 seconds.

G) Find the velocity when the acceleration is 0 m/s^2 .

Answer 1

The position of a particle moving along the 𝑥-axis is given by 𝑠(𝑡) = 5 + 3𝑡^2 − 2𝑡^3 meters for 𝑡 ≥ 0 seconds.

A) Find the initial position, 𝑡 = 0, of the particle. s(0) = 5 + 3(0)² - 2(0)³ = 5 meters

B) Find the displacement for 1 ≤ 𝑡 ≤ 3. s(3) - s(1) = -22 - 6 = -28 meters

C) Find the average velocity for 1 ≤ 𝑡 ≤ 3. [ s(3) - s(1) ] / (3 - 1) = -28/2 = -14 m/s

D) Find the velocity at 𝑡 =2 seconds. Find the speed of the particle at 𝑡 = 2 seconds. Is the ball moving forward or backward at 𝑡 = 2 seconds? Justify. v = s '(t) = 6t - 6t² v(2) = -12 m/s speed = 12 m/s moving backwards b/c of the velocity being negative. (speed is the magnitude of the velocity)

E) How long does it take for the velocity to reach −36 m/s? -36 = 6t - 6t² -6 = t - t² so t² - t -6 = 0 factor this into (t-3)(t+2)=0 so t = 3 or t = -2 Choose t = 3 seconds

F) Find the acceleration at 𝑡 = 2 seconds. accel is v '(t) a(t) = 6 - 12t a(2) = 6 - 12(2) = 6 - 24 = -18 m/s²

G) Find the velocity when the acceleration is 0 m/s^2 . 0 = 6 - 12 t 12t = 6 t = 2, v(2) = -12 m/s