I'm having difficulty understanding this concept, and I was wondering if anyone could help me with this multi-step problem. I would greatly appreciate any help.
The position (x - coordinate) of a particle moving on the line y = 2 is given by x(t) = 2t^3 -13t^2 + 22t - 5 where t is time in seconds.
a) Describe the motion of the particle for t is greater than or equal to zero.
b) When does the particle speed up? slow down?
c) When does the particle change direction?
d) When is the particle at rest?
e) Describe the velocity and speed of the particle.
f) When is the particle at the point (5,2)?
I believe to find speed, one has to take the first derivative, and to find the change of direction one has to find the second derivative, but I am not even sure of this. Thank you so much.
That being said:
Originally Posted by turtle
a) To describe the motion, the simplest thing to do is to graph the x(t) function. I'll do a Calculus based analysis of the motion, which pretty much answers all the subsequent questions.
x(t) = 2t^3 -13t^2 + 22t - 5
x'(t) = 6t^2 - 26t + 22
x"(t) = 12t - 26
The first thing to note is that at t = 0 s the particle is at x = -5 (m, I presume), so the particle is not starting from our origin. Also x'(0) = 22 so the particle is starting with a positive velocity, and since x"(0) = -26, this velocity is getting smaller.
b) x"(t) = 12t - 26
Set 12t - 26 = 0 ==> t = 13/6 s.
This is the time when the velocity switches over from being positive and goes negative. So the velocity is positive for [0, 13/6) and negative for (13/6, infinity).
c) To get a changing direction we need to see where the velocity is 0:
x'(t) = 6t^2 - 26t + 22
0 = 6t^2 - 26t + 22
0 = 3t^2 - 13t + 11
So t = (13 (+/-) sqrt(37))/6 s is when the velocity of the particle is 0. Now, we are taking t >=0 s, so we discard the negative solution and get t = (13 + sqrt(37))/6 s. So initially the velocity is positive and at this time the velocity switches and goes negative. Thus the particle changes direction at t = (13 + sqrt(37))/6 s.
(Note carefully I have not exactly proven a sign change exists in either b) or c). I leave that detail to you.)
d) The particle never "rests" as the velocity is always changing. However I suspect that the question is asking where v = 0. We already know this from the answer to c). The particle momentarily stops at t = (13 + sqrt(37))/6 s.
e) I'll leave the details of this to you. You have the velocity function x'(t) = 6t^2 - 26t + 22, which you can graph, find the zeros of, etc. The speed function is the magnitude of the velocity function: s(t) = |x'(t)| = |6t^2 - 26t + 22|
f) This problem can be solved numerically, but I wonder if we aren't looking for when x = -5? This has a much simpler solution.
Anyway, we are asking when x = 5. So solve
5 = 2t^3 -13t^2 + 22t - 5
0 = 2t^3 - 13t^2 + 22t - 10
I get t = 0.744721 s and and t = 1.62589 s and t = 4.12939 s as solutions.
A toy rocket vertically in such a way that t seconds after lift-off, it is
h(t) = 1/2t^2 + 30t feet above ground.
a) How high is the rocket after 40 seconds?
b) What is the average velocity of the rocket over the first 40 seconds of flight (between t = 0 and t = 40)?
c)What is the (instantaneous) velocity of the rocket at lift-off (t = 0)? What is the velocity after 40 seconds?(Speechless)
please post your questions in a new thread
Originally Posted by cclia