posted by Lauren .
At time t = 0s, a puck is sliding on a horizontal table with a velocity 3.00 m/s, 65.0 above the +x axis. As the puck slides, a constant acceleration acts on it that was the following components: ax = 0.460m/s2 and ay = 0.980m/s2. what is the velocity of the puck at time t = 1.50s?
If the x and y axes are on the horizontal table, how can the puck be "above" the x axis and sliding at the same time?
Perhaps the 65.0 refers to the direction angle in degrees, measured from the x axis.
In that case, at t = 1.50 s,
Vx = 3.00 cos 65 + 0.460 * 1.50
Vy = 3.00 sin 65 + 0.980 * 1.50
Compute both and then take the square root of the sum of the squares for the magnitude of the velocity (speed)
the 65 does not represent the degrees. . .that is why i am so confused on this problem