s0 = 28.7 @ 95.0° = -2.501,28.591
v0 = 4.85 @ 40.0° = 3.715,3.119
a = 1.94 @ 200° = -1.823,-0.664
Now, we just plug in our usual equations:
v = v0 + at
s = s0 + v0*t + 1/2 at^2
and then convert back to polar form as needed.
A snowmobile is originally at the point with position vector 28.7 m at 95.0° counterclockwise from the x axis, moving with velocity 4.85 m/s at 40.0°. It moves with constant acceleration 1.94 m/s2 at 200°. After 5.00 s have elapsed, find the following:
(a)Its velocity vector
(b)Its position vector
1 answer