Kinematics?
Answer:
The x,y,z, components of the position vector are found by integrating each of the x,y,z components of the velocity vector. The constants of integration are determined by the position of the particle at t = 0.
dx(t)/dt = 3t
x(t) = (3*t^2)/2 + 1
dy(t)/dt = -4
y(t) = -4*t + 0
dz(t)/dt = t^2
z(t) = (t^3)/3 + 1
So the position vector is:
r(t) = [(3*t^2)/2 + 1]i +[-4*t]j + [(t^3)/3 + 1]k
at t = 2
r(2) = 7i - 8j + 11k/3
The acceleration vector is found by taking the derivative of each of the components of the velocity vector:
d^2x(t)/dt^2 = 3
d^2y(t)/dt^2 = 0
d^2z(t)/dt^2 = 2t
a(t) = [3i + 2tk]
The magnitude of the acceleration at any time is given by the square root of the sum of the squares of the x, y, and z components of the acceleration:
|a(t)| = sqrt(9 + 4t^2)
At t = 2, the magnitude of the acceleration is therefore:
|a(2)| = sqrt(9 + 16) = sqrt(25) = 5
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