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 = +2.2 m/s and ax = +2.5 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +7.3 m/s and ay = -6.2 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.
vx = 2.2 + (2.5 * 0.50) = 3.5 m/s
vy = 7.3 - (6.2 * 0.50) = 4.2 m/s
(a) v^2 = vx^2 + vy^2 = 3.5^2 + 4.2^2
(b) tan(θ) = vy / vx = 4.2 / 3.5
To find the magnitude and direction of the puck's velocity at t = 0.50 s, we can use the equations of motion.
We can find the x and y components of the velocity at t = 0.50 s using the equations:
vx = v0x + ax * t
vy = v0y + ay * t
Given:
v0x = +2.2 m/s
ax = +2.5 m/s^2
v0y = +7.3 m/s
ay = -6.2 m/s^2
t = 0.50 s
(a) The magnitude of the velocity is given by the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
Let's calculate the x and y components of the velocity first:
vx = v0x + ax * t
= 2.2 m/s + 2.5 m/s^2 * 0.50 s
= 2.2 m/s + 1.25 m/s
= 3.45 m/s
vy = v0y + ay * t
= 7.3 m/s + (-6.2 m/s^2) * 0.50 s
= 7.3 m/s + (-3.1 m/s)
= 4.2 m/s
Now we can calculate the magnitude of the velocity:
v = sqrt(vx^2 + vy^2)
= sqrt((3.45 m/s)^2 + (4.2 m/s)^2)
= sqrt(11.9025 m^2/s^2 + 17.64 m^2/s^2)
= sqrt(29.5425 m^2/s^2)
≈ 5.438 m/s
Therefore, the magnitude of the puck's velocity at t = 0.50 s is approximately 5.438 m/s.
(b) The direction of the velocity can be found using trigonometry. The angle θ can be calculated using the inverse tangent function:
θ = atan(vy / vx)
Let's calculate θ:
θ = atan(4.2 m/s / 3.45 m/s)
= atan(1.217391304)
≈ 0.891 radians
To specify the direction relative to the +x axis, we convert the angle to degrees:
θ_degrees = θ * (180 / π)
= 0.891 radians * (180 / π)
≈ 51.05 degrees
Therefore, the direction of the puck's velocity at t = 0.50 s relative to the +x axis is approximately 51.05 degrees.