Question

The acceleration of a particle is defined by the relation a = 9 − 3t2, where a and t are expressed in ft/s2 and seconds, respectively. The particle starts at t = 0 with v = 0 and x = 5 ft. Determine (a) the time when the velocity is again zero, (b) the position and velocity when t = 4 s, (c) the total distance travelled by the particle from t = 0 to t = 4 s.

Answer 1

(a) To find the time when the velocity is again zero, we set the velocity equal to 0 and solve for t:

0 = v0 + at = 9t - 3t^3

Solving for t, we find that t = ±1.

(b) To find the position and velocity when t = 4 s, we first need to find the velocity equation by integrating the acceleration equation:

v = ∫a = ∫(9 - 3t^2)dt = 9t - t^3 + C

Using the initial condition v(0) = 0, we find C = 0 and

v(t) = 9t - t^3

Next, we integrate v(t) to find the position equation:

x = ∫v = ∫(9t - t^3)dt = 3t^2 - t^4 + C

Using the initial condition x(0) = 5, we find C = 5 and

x(t) = 3t^2 - t^4 + 5

Evaluating x(t) and v(t) at t = 4 s, we have:

x(4) = 3(4^2) - 4^4 + 5 = 51

v(4) = 9(4) - 4^3 = -36

(c) To find the total distance travelled by the particle from t = 0 to t = 4 s, we need to find the distance between the initial and final positions:

d = x(4) - x(0) = 51 - 5 = 46 ft

So the total distance travelled by the particle from t = 0 to t = 4 s is 46 ft.