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 = +3.9 m/s and ax = +1.2 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +4.7 m/s and ay = -1.8 m/s2. Find (a) the magnitude v and (b) the direction θ of the puck's velocity at a time of t = 0.50 s. Specify the direction relative to the +x axis.

To find the magnitude v of the puck's velocity at t = 0.50 s, we can use the equations of motion. The magnitude of velocity, v, can be found using the Pythagorean theorem:

v = √(vx^2 + vy^2)

where vx and vy are the x and y components of the velocity respectively.

Given:
vx = v0x + ax*t
= 3.9 m/s + 1.2 m/s^2 * 0.50 s
= 3.9 m/s + 0.6 m/s
= 4.5 m/s

vy = v0y + ay*t
= 4.7 m/s - 1.8 m/s^2 * 0.50 s
= 4.7 m/s - 0.9 m/s
= 3.8 m/s

Now, we can substitute these values back into the equation for v:

v = √(4.5^2 + 3.8^2)
= √(20.25 + 14.44)
= √(34.69)
≈ 5.89 m/s

Therefore, the magnitude of the puck's velocity at t = 0.50 s is approximately 5.89 m/s.

To find the direction θ of the puck's velocity relative to the +x axis, we can use the tangent function:

θ = arctan(vy / vx)

θ = arctan(3.8 m/s / 4.5 m/s)
≈ arctan(0.844)
≈ 40.23°

Therefore, the direction of the puck's velocity at t = 0.50 s relative to the +x axis is approximately 40.23°.