Question

The position, x metres, of a particle travelling in a straight line from a fixed origin at time t seconds is given by x (t) = 3t^3 - 9t^2 + 12t - 2, 0 ≤ t ≤ 3.
Determine, correct to 2 decimal places, the time(s) when the numerical component of the particle’s velocity and the numerical component of the acceleration are equal.

Answer 1

To find the velocity and acceleration of the particle, we will need to take the derivative of x(t) with respect to t.

Velocity v(t):
v(t) = dx/dt = d/dt (3t^3 - 9t^2 + 12t - 2)
v(t) = 9t^2 - 18t + 12

Acceleration a(t):
a(t) = dv/dt = d/dt (9t^2 - 18t + 12)
a(t) = 18t - 18

Now, we will set the numerical components of velocity and acceleration equal to each other and solve for t:

9t^2 - 18t + 12 = 18t - 18
9t^2 - 36t + 30 = 0

Using the quadratic formula, we have:
t = (36 ± sqrt((-36)^2 - 4*9*30)) / (2*9)
t = (36 ± sqrt(1296 - 1080)) / 18
t = (36 ± sqrt(216)) / 18
t = (36 ± 14.7) / 18

Therefore, t = (36 + 14.7) / 18 or t = (36 - 14.7) / 18
t = 2.04 or t = 1.31

Therefore, the time(s) when the numerical component of the particle’s velocity and the numerical component of the acceleration are equal are approximately 1.31 seconds and 2.04 seconds.