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 = +0.8 m/s and ax = +0.9 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +1.6 m/s and ay = -2.4 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.
(a) Finding the magnitude of velocity (v):
The magnitude of velocity can be calculated using the Pythagorean theorem, which states that the magnitude of a vector is the square root of the sum of the squares of its components.
v = √(v_x^2 + v_y^2)
Given:
v0x = +0.8 m/s (initial x-component velocity)
v0y = +1.6 m/s (initial y-component velocity)
Using the equations of motion, we can find the x and y-components of velocity at time t = 0.50 s:
v_x = v0x + a_x * t
v_y = v0y + a_y * t
Substituting the given values:
v_x = 0.8 m/s + 0.9 m/s^2 * 0.50 s
v_y = 1.6 m/s + (-2.4 m/s^2) * 0.50 s
Calculate:
v_x = 0.8 m/s + 0.45 m/s
v_y = 1.6 m/s + (-1.2 m/s)
v_x = 1.25 m/s
v_y = 0.4 m/s
Now, calculate the magnitude (v):
v = √(v_x^2 + v_y^2)
v = √(1.25^2 + 0.4^2)
v ≈ 1.32 m/s
Therefore, the magnitude of the puck's velocity at t = 0.50 s is approximately 1.32 m/s.
(b) Finding the direction (θ) of the puck's velocity:
The direction of the puck's velocity can be calculated using the inverse tangent function (arctan).
θ = arctan(v_y / v_x)
Substituting the calculated values:
θ = arctan(0.4 m/s / 1.25 m/s)
Calculate:
θ ≈ arctan(0.32)
θ ≈ 16.27°
Therefore, the direction of the puck's velocity at t = 0.50 s relative to the +x axis is approximately 16.27°.