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 = +5.3 m/s and ax = +2.6 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +6.2 m/s and ay = -3.3 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 and direction of the puck's velocity at a time of t = 0.50 s, we can use the equations of motion. We'll break down the puck's velocity into its x and y components, and then calculate the magnitude and direction based on these components.

Step 1: Calculate the x component of the velocity at t = 0.50 s:
The x component of the initial velocity, v0x = +5.3 m/s.
The acceleration in the x direction, ax = +2.6 m/s^2.
The time, t = 0.50 s.

Using the equation of motion for the x component of velocity:
vx = v0x + ax * t

Substituting the values, we have:
vx = 5.3 m/s + 2.6 m/s^2 * 0.50 s
= 5.3 m/s + 1.3 m/s
= 6.6 m/s

Therefore, the x component of the velocity at t = 0.50 s is 6.6 m/s.

Step 2: Calculate the y component of the velocity at t = 0.50 s:
The y component of the initial velocity, v0y = +6.2 m/s.
The acceleration in the y direction, ay = -3.3 m/s^2.
The time, t = 0.50 s.

Using the equation of motion for the y component of velocity:
vy = v0y + ay * t

Substituting the values, we have:
vy = 6.2 m/s + (-3.3 m/s^2) * 0.50 s
= 6.2 m/s + (-1.65 m/s)
= 4.55 m/s

Therefore, the y component of the velocity at t = 0.50 s is 4.55 m/s.

Step 3: Calculate the magnitude of the velocity at t = 0.50 s:
The magnitude of the velocity, v, can be found using the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)

Substituting the values, we have:
v = sqrt((6.6 m/s)^2 + (4.55 m/s)^2)
≈ sqrt(43.56 m^2/s^2 + 20.70 m^2/s^2)
≈ sqrt(64.26 m^2/s^2)
≈ 8.02 m/s

Therefore, the magnitude of the puck's velocity at t = 0.50 s is approximately 8.02 m/s.

Step 4: Calculate the direction of the velocity at t = 0.50 s:
The direction of the velocity is given by the angle θ relative to the +x axis. We can use trigonometry to find this angle.

Using the equation:
θ = arctan(vy / vx)

Substituting the values, we have:
θ = arctan(4.55 m/s / 6.6 m/s)
= arctan(0.690)
≈ 34.35°

Therefore, the direction of the puck's velocity at t = 0.50 s, relative to the +x axis, is approximately 34.35°.