Question

A particle at rest leaves the origin with its velocity increasing with time according to v(t) = 3.2t m/s. At 5.0 s, the particle’s velocity starts decreasing according to [16.0 – 1.5(t – 5.0)] m/s. This decrease continues until t = 11.0 s, after which the particle’s velocity remains constant at 7.0 m/s. (a) What is the acceleration of the particle as a function of time? (b) What is the position of the particle at t = 2.0 s, t = 7.0 s, and t = 12.0 s?

a) Option a: a(t) = 3.2 m/s^2 for 0 ≤ t ≤ 5, a(t) = -1.5 m/s^2 for 5 < t < 11, a(t) = 0 for t ≥ 11; Option b: x(2.0 s) = 6.4 m, x(7.0 s) = 42.7 m, x(12.0 s) = 86.7 m
b) Option a: a(t) = 3.2 m/s^2 for 0 ≤ t ≤ 5, a(t) = -1.5 m/s^2 for 5 < t < 11, a(t) = 7.0 m/s^2 for t ≥ 11; Option b: x(2.0 s) = 6.4 m, x(7.0 s) = 37.0 m, x(12.0 s) = 94.0 m
c) Option a: a(t) = 3.2 m/s^2 for 0 ≤ t ≤ 5, a(t) = -1.5 m/s^2 for 5 < t < 11, a(t) = 1.5 m/s^2 for t ≥ 11; Option b: x(2.0 s) = 3.2 m, x(7.0 s) = 40.3 m, x(12.0 s) = 99.7 m
d) Option a: a(t) = 3.2 m/s^2 for 0 ≤ t ≤ 5, a(t) = 1.5 m/s^2 for 5 < t < 11, a(t) = 0 for t ≥ 11; Option b: x(2.0 s) = 3.2 m, x(7.0 s) = 48.5 m, x(12.0 s) = 91.0 m

Answer 1

The acceleration of the particle as a function of time is 3.2 m/s^2 from t = 0 to t = 5, -1.5 m/s^2 from t = 5 to t = 11, and 0 after t = 11. The position of the particle at different times is x(2.0 s) = 6.4 m, x(7.0 s) = 42.7 m, and x(12.0 s) = 86.7 m.

Step-by-step explanation:

a) The acceleration of the particle as a function of time can be determined based on the given information. From t = 0 to t = 5, the acceleration is constant at a(t) = 3.2 m/s^2. From t = 5 to t = 11, the acceleration is a(t) = -1.5 m/s^2. After t = 11, the acceleration is 0, meaning it remains constant at that point.

b) The position of the particle at various times can be calculated by integrating the velocity function. The position at t = 2.0 s is x(2.0 s) = 6.4 m, at t = 7.0 s is x(7.0 s) = 42.7 m, and at t = 12.0 s is x(12.0 s) = 86.7 m.