A particle moves on a vertical line so that its coordinate at time t is y = t^3

Question

A particle moves on a vertical line so that its coordinate at time t is y = t^3 - 12t + 3, t greaterthanorequalto 0. Find the velocity and acceleration functions. When is the particle moving upward and when is it moving downward? Find the distance that the particle travels in the time interval 0 lessthanorequalto t lessthanorequalto 3. Graph the position, velocity, and acceleration functions for 0 lessthanorequalto t lessthanorequalto 3. When is the particle speeding up? When is it slowing down?

Explanation / Answer

Now, Find the velocity function of the particle.
v(t) = s'(t) =3t^2 -12
Find the acceleration function of the particle.
a(t) = v'(t) = 6t
This is asking for the time(s) at which v(t) = 0. From part (a) we solve and get
3t^2 12 = 0, so
that 3(t + 2)(t 2) = 0, which give t = 2 or t = 2. Since we are told t 0, we get that t = 2 seconds is the
hat time interval(s) is the particle moving upward (i.e., when is the particle’s position increasing)
we know v(t) = 0 when t = 2
sec. Plugging in any value between 0 and 2 (e.g. t = 1) shows that v(t) < 0 on this interval. Similarly, we see
that v(t) > 0 when t > 2. So the particle is moving downward initially on the interval 0 t < 2 sec, and then
upward when t > 2 sec.
(c) Find the distance that the particle travels in the time interval ,0 t 3.
We need to analyze separately the two intervals indicated On the interval 0 t < 2 the
position of the particle changes from s(0) to s(2), namely position 03 12 · 0 + 3 to position 23 12 · 2 + 3, or
3 to 13. This is 16 meters. On the interval 2 t 3 the position of the particle changes from s (2) to s(3),
namely position 23 12 · 2 + 3 to 33 12 · 3 + 3, or 13 to 6. This is a change of 7 meters. So the total distance
travelled in the time interval 0 t 3 is 16 + 7 = 23 meters.

e) To find the times when the particle is accelerating/decelerating,
dv /dt = a = 6t
As you can see here, the particle is accelerating whenever t is greater than 0, and decelerating whenever t is less than 0.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.