At two points (1 and 2) in the motion of a particle, the instantaneous velocity?

Question

At two points (1 and 2) in the motion of a particle, the instantaneous velocity
vectors have equal magnitudes of 20 m/s. At point 1, the velocity vector ~v1 points in the
positive x-direction. At point 2, the velocity vector ~v2 is at angle −45 to the positive
x-direction.
(a) Resolve ~v1 and ~v2 into rectangular scalar components. Do not forget the physical units.
(b) Take the difference of the two velocity vectors, ~v = ~v2 − ~v. Express this difference
in rectangular components.
(c) Divide the difference you found in part (b) by the time t = 6 s it took the particle
to travel between the two points. This gives ~v/t, the average acceleration ~aavg of the
particle over the interval between point 1 and 2 on the trajectory. Give the magnitude and
direction (angle with respect to the x-axis) of this average acceleration.

Answer 1

Given a vector (r, Θ) you get the rectangular components with
x = r*cos(Θ)
y = r*sin(Θ)
and remember that Θ = 0 'points' to the right and increasing Θ is revolution counterclockwise ☺

The rest of it you should be able to do if you were paying any attention at all to the one dimensional equations of motion. If not, it's time to go back and learn them.


Doug