A ball rolling with initial velocity of 40m/s(W) undergoes an acceleration of 5.0m/s^2(N) for a period of 6.0seconds.

Question

A ball rolling with initial velocity of 40m/s(W) undergoes an acceleration of 5.0m/s^2(N) for a period of 6.0seconds.
(A) what is the final velocity of the ball?
(B) what is the displacement of the ball in the 6.0 sec

Henry · Answer

X = -40 m/s.
Y = 5m/s^2 * 6s. = 30 m/s.

A. Tan Ar = Y/X = 30/-40 = -0.750
Ar = -36.9o = Reference angle.
A = -36.9 + 180 = 143.1o CCW = 53.1o W.
of N.

V = X/Cos A = -40/Cos143.1 = 50 m/s @
143.1o CCW or 53.1o W. of N.