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 = +5.2 m/s and ax = +5.9 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +2.6 m/s and ay = -1.3 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.
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PM.
To find the magnitude and direction of the puck's velocity at t = 0.50 seconds, we can use the equations of motion for projectile motion.
Step 1: Find the x-component of velocity (vx) at t = 0.50s.
The x-component of the initial velocity is given as v0x = +5.2 m/s.
The x-component of acceleration is ax = +5.9 m/s^2.
Using the equation: vx = v0x + ax*t, substitute the given values:
vx = 5.2 m/s + (5.9 m/s^2)*(0.50 s)
vx = 5.2 m/s + 2.95 m/s
vx = 8.15 m/s
Step 2: Find the y-component of velocity (vy) at t = 0.50s.
The y-component of the initial velocity is given as v0y = +2.6 m/s.
The y-component of acceleration is ay = -1.3 m/s^2.
Using the equation: vy = v0y + ay*t, substitute the given values:
vy = 2.6 m/s + (-1.3 m/s^2)*(0.50 s)
vy = 2.6 m/s - 0.65 m/s
vy = 1.95 m/s
Step 3: Find the magnitude of the velocity (v) at t = 0.50s.
The magnitude of the velocity v is given by the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
v = sqrt((8.15 m/s)^2 + (1.95 m/s)^2)
v = sqrt(66.5225 m^2/s^2 + 3.8025 m^2/s^2)
v = sqrt(70.325 m^2/s^2)
v ≈ 8.38 m/s
Step 4: Find the direction θ of the velocity at t = 0.50s.
The direction θ can be found using trigonometry.
θ = arctan(vy/vx)
θ = arctan(1.95 m/s / 8.15 m/s)
θ ≈ 0.238 radians
To specify the direction relative to the +x axis, we can convert the angle to degrees:
θ ≈ 0.238 radians * (180°/π radians)
θ ≈ 13.64°
Therefore, at t = 0.50 seconds, the magnitude of the puck's velocity is approximately 8.38 m/s, and the direction relative to the +x axis is approximately 13.64°.