A particle's position is given by
z(t) = −(8.10 m/s^2)t^2k
for
t ≥ 0.
(Express your answer in vector form.)
(a) Find the particle's velocity at
t = 3.00 s and t = 6.00 s.
just do the usual derivative:
v(t) = z'(t) = -16.2tk
and evaluate at t=3 and t=6
works the same if there are i and j components as well.
To find the particle's velocity at t = 3.00 s and t = 6.00 s, we need to take the derivative of the position function with respect to time.
Given that the position function is z(t) = -(8.10 m/s^2)t^2k, where k is the unit vector in the z-direction, we can differentiate it to find the velocity function.
The derivative of a position function with respect to time gives the velocity function. Since the function only has a z-component, the velocity vector will also only have a z-component.
Now let's find the derivative of z(t):
dz/dt = d/dt(-(8.10 m/s^2)t^2) * k
The derivative of -(8.10 m/s^2)t^2 with respect to t is:
d/dt(-(8.10 m/s^2)t^2) = -(2)(8.10 m/s^2)t = -16.20 m/s^2t
Therefore, the velocity function is v(t) = -16.20 m/s^2t * k.
Now we can substitute the given values of t to find the velocity at those times:
a) At t = 3.00 s:
v(3.00) = -16.20 m/s^2 * (3.00 s) * k
= -48.60 m/s * k
So the velocity at t = 3.00 s is -48.60 m/s in the z-direction.
b) At t = 6.00 s:
v(6.00) = -16.20 m/s^2 * (6.00 s) * k
= -97.20 m/s * k
So the velocity at t = 6.00 s is -97.20 m/s in the z-direction.
Therefore, the particle's velocity at t = 3.00 s is -48.60 m/s in the z-direction, and at t = 6.00 s is -97.20 m/s in the z-direction.