A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0 s, the x components of the puck's initial velocity and acceleration are v0x = +3.5 m/s and ax = +5.8 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +2.8 m/s and ay = -2.0 m/s2. Find (a) the magnitude v and (b) the direction è of the puck's velocity at a time of t = 0.50 s. Specify the direction relative to the +x axis.
To solve this problem, we can use the equations of motion to find the magnitude and direction of the velocity of the puck at time t = 0.50 s.
First, let’s find the x-component of the velocity (vx) using the equation:
vx = v0x + ax * t
Given:
v0x = +3.5 m/s (initial velocity in x-direction)
ax = +5.8 m/s^2 (acceleration in x-direction)
t = 0.50 s (time)
Substituting the given values into the equation, we have:
vx = 3.5 m/s + 5.8 m/s^2 * 0.50 s
vx = 3.5 m/s + 2.9 m/s
vx = 6.4 m/s
Next, let’s find the y-component of the velocity (vy) using the equation:
vy = v0y + ay * t
Given:
v0y = +2.8 m/s (initial velocity in y-direction)
ay = -2.0 m/s^2 (acceleration in y-direction)
t = 0.50 s (time)
Substituting the given values into the equation, we have:
vy = 2.8 m/s + (-2.0 m/s^2) * 0.50 s
vy = 2.8 m/s - 1.0 m/s
vy = 1.8 m/s
To find the magnitude of the velocity (v), we can use the Pythagorean theorem:
v = √(vx^2 + vy^2)
Substituting the calculated values for vx and vy, we have:
v = √(6.4 m/s)^2 + (1.8 m/s)^2
v ≈ √(40.96 m^2/s^2 + 3.24 m^2/s^2)
v ≈ √44.20 m^2/s^2
v ≈ 6.65 m/s
Finally, to find the direction è of the velocity relative to the +x axis, we can use the inverse tangent function:
è = arctan(vy / vx)
Substituting the calculated values for vy and vx, we have:
è = arctan(1.8 m/s / 6.4 m/s)
è ≈ arctan(0.28125)
è ≈ 15.69 degrees
Therefore, the magnitude of the velocity (v) at t = 0.50 s is approximately 6.65 m/s, and the direction è of the velocity relative to the +x axis is approximately 15.69 degrees.