Answer:
For the particle to move to the right, 0 ≤ t ≤ a
Step-by-step explanation:
The particle under motion on the x-axis given by the equation x(t) = 12(a−t)² for t ≥ 0, where a is a positive constant.
To find the values of t for which the particle is moving to the right, this implies that the particle must move along the positive x-axis. This implies that x(t) ≥ 0.
So, to find the value of t for which the particle moves to the right, we solve the inequality x(t) ≥ 0 ⇒ 12(a−t)² ≥ 0
12(a−t)² ≥ 0
(a−t)² ≥ 0
taking square root of both sides, we have
√(a−t)² ≥ √0
(a−t) ≥ 0
a ≥ t ⇒ t ≤ a
So, for the particle to move to the right, 0 ≤ t ≤ a.
Using derivatives, it is found that the particle is moving to the right for [tex]t > a[/tex], that is, values of t in the interval [tex](a, \infty)[/tex].
A particle is moving to the right if it's velocity is positive.
The position of the particle is given by:
[tex]x(t) = 12(a - t)^2[/tex]
The velocity is the derivative of the position, hence:
[tex]v(t) = -24(a - t)[/tex]
If will be positive if:
[tex]-24(a - t) > 0[/tex]
[tex]-(a - t) > 0[/tex]
[tex]-a + t > 0[/tex]
[tex]t > a[/tex]
A similar problem is given at https://brainly.com/question/14516604
