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 = +2.5 m/s and ax = +5.2 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +2.8 m/s and ay = -2.3 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 v of the puck's velocity at a time of t = 0.50 s, we can use the formula for velocity:
v = √(v_x^2 + v_y^2)
where v_x and v_y are the x and y components of the velocity, respectively.
Given v_0x = +2.5 m/s and v_0y = +2.8 m/s, we can calculate v_x and v_y at time t = 0.50 s using the equations of motion:
v_x = v_0x + a_x * t
v_y = v_0y + a_y * t
Plugging in the given values, we get:
v_x = 2.5 m/s + 5.2 m/s^2 * 0.50 s = 2.5 m/s + 2.6 m/s = 5.1 m/s
v_y = 2.8 m/s + (-2.3 m/s^2) * 0.50 s = 2.8 m/s - 1.15 m/s = 1.65 m/s
Now, we can calculate the magnitude v:
v = √((5.1 m/s)^2 + (1.65 m/s)^2) = √(26.01 m^2/s^2 + 2.7225 m^2/s^2)
v = √28.7325 m^2/s^2 ≈ 5.36 m/s
Next, to find the direction θ of the puck's velocity, we can use the tangent of θ:
θ = arctan(v_y / v_x)
Plugging in the values we calculated, we get:
θ = arctan(1.65 m/s / 5.1 m/s) ≈ arctan(0.323)
θ ≈ 18.08°
Therefore, the magnitude v of the puck's velocity at t = 0.50 s is approximately 5.36 m/s, and the direction θ is approximately 18.08° relative to the +x axis.