Question
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?
Answers
drwls
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)
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)
Lauren
the 65 does not represent the degrees. . .that is why i am so confused on this problem