Question

A particle starts moving from rest with an acceleration of a(t) = (6t - 4) ft/s2, where t is in seconds. what is the particle's position at t = 4 s??

Answer 1

The position of a particle that starts from rest with the acceleration a(t) = (6t - 4) ft/s² at t = 4 seconds is found by integrating the acceleration to get velocity and then position. After integration, the position at t = 4 s is 32 feet from the origin.

Step-by-step explanation:

The student asks about a particle that starts from rest with a given acceleration function, a(t) = (6t - 4) ft/s², and we need to find the particle's position at t = 4 s. To find the position, we integrate the acceleration function to get the velocity function, v(t), and then integrate v(t) to get the position function, x(t). Since the particle starts from rest, v(0) = 0, and x(0) would be the initial position, which we can consider to be zero if not given.

1. Integrate acceleration to find velocity: ∫ a(t) dt = ∫ (6t - 4) dt = 3t² - 4t + C. Since v(0) = 0, C = 0, and thus v(t) = 3t² - 4t.
2. Integrate velocity to find position: ∫ v(t) dt = ∫ (3t² - 4t) dt = t³ - 2t² + D. As x(0) = 0, D = 0, and thus x(t) = t³ - 2t².
3. Substitute t = 4 s into x(t) to find the position: x(4) = 4³ - 2(4)² = 64 - 32 = 32 feet.

Therefore, the position of the particle at t = 4 s is 32 feet from the starting point.