Question

Question C) needs to be answered, please help (physics)

Answer 1

(a) Differentiate the position vector to get the velocity vector:

r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k

v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j

(b) The velocity at t = 2.00 s is

v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j

(c) Compute the electron's position at t = 2.00 s:

r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k

The electron's distance from the origin at t = 2.00 is the magnitude of this vector:

||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that

tan(θ) = (-16.0 m/s) / (3.00 m/s)   ==>   θ ≈ -79.4º

or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.