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.

1 answer

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.
Similar Questions
  1. In the following math question:The position, x metres, of a particle travelling in a straight line from a fixed origin at time t
    1. answers icon 1 answer
  2. A particle travelling along a straight line starts with avelociry ofu metres per second and its speed increases per second by rl
    1. answers icon 1 answer
    1. answers icon 2 answers
    1. answers icon 2 answers
more similar questions