A particle of mass 2.0 kg has position given by the vector r = (t^4 -3t^3) i + (t^3 - 9t) j + (2t -6) k
(where i, j, and k are the unit direction vectors and r is given in meters and t in seconds. At what time in seconds is the particle located at the origin?
For this I get t = 3(seconds).
It's the second part I want you to check.
1. what is the magnitude of the particle's velocity (in m/s) when the particle is at the origin?
2. For the particle in question 1, what is the x/y/z-component of the net force (in Newtons) acting on it when it is at the origin? Do not write Newtons as part of your answer.
V = Dr / Dt
The Attempt at a Solution
1. Derive each component individually. Got:
(4t^3-9t^2)i, (3t^2 - 9)j, (2)k
Plug in 3 to get. 27 + 18 + 2 = Sqrt 47 = 6.86 m/s while at origin.
2. Derived each component again to get the acceleration eq.
I got (12t^2 - 18t)i, (6t)j, and (0)k
Then did F = MA for each component.
2.0(12(3)^2 - 18(3)) = 108 for x component.
2.0(6(3)) = 36 for the y component.
2.0(0) = 0 for the z component.
Did I do this correctly?
Thanks for helping :D.