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 = +3.2 m/s and ax = +1.9 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +8.1 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.
X = Xo + a*t
X = 3.2 + 1.9*0.5 = 4.15 m/s^2
Y = Yo + a*t
Y = 8.1 - 6.2*0.5 = 5 m/s.
a. V^2 = X^2 + Y^2
V^2 = (4.15)^2 + 5^2 = 42.2
V = 6.5 m/s.
b. tan A = Y/X = 5/4.15 = 1.20482
A = 50.31o = Direction.
Correction: X = 4.15 m/s.
To find the magnitude and direction of the puck's velocity at a time of t = 0.50 s, we can use the following steps:
Step 1: Calculate 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 = +3.2 m/s (initial x velocity)
ax = +1.9 m/s^2 (x acceleration)
v0y = +8.1 m/s (initial y velocity)
ay = -6.2 m/s^2 (y acceleration)
t = 0.50 s (time)
Plugging in the values:
vx = 3.2 m/s + 1.9 m/s^2 * 0.50 s = 4.1 m/s
vy = 8.1 m/s + (-6.2 m/s^2) * 0.50 s = 5.0 m/s
Step 2: Calculate the magnitude of the velocity using the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
Plugging in the values:
v = sqrt(4.1 m/s^2)^2 + (5.0 m/s^2)^2 = sqrt(16.81 m^2/s^4 + 25.00 m^2/s^4)
v = sqrt(41.81 m^2/s^4) = 6.47 m/s
Step 3: Calculate the direction of the velocity relative to the +x axis using the arctan function:
θ = arctan(vy / vx)
Plugging in the values:
θ = arctan(5.0 m/s / 4.1 m/s) = 51.79 degrees
(a) The magnitude of the puck's velocity at t = 0.50 s is 6.47 m/s.
(b) The direction of the puck's velocity at t = 0.50 s, relative to the +x axis, is 51.79 degrees.
To find the magnitude (v) and direction (θ) of the puck's velocity at t = 0.50 s, we can use the equations of motion.
(a) Finding the magnitude (v) of the velocity:
To find the magnitude of the velocity, we can use the Pythagorean theorem. The magnitude of the velocity (v) can be found using the equation:
v = √(vx^2 + vy^2)
Given:
vx = 3.2 m/s (initial x-component of velocity)
vy = 8.1 m/s (initial y-component of velocity)
Using the equation, we have:
v = √(3.2^2 + 8.1^2)
v = √(10.24 + 65.61)
v = √75.85
v ≈ 8.71 m/s
Therefore, the magnitude of the puck's velocity at t = 0.50 s is approximately 8.71 m/s.
(b) Finding the direction (θ) of the velocity:
To find the direction of the velocity, we can use trigonometry. The direction (θ) can be found using the equation:
θ = arctan(vy / vx)
Given:
vx = 3.2 m/s (initial x-component of velocity)
vy = 8.1 m/s (initial y-component of velocity)
Using the equation, we have:
θ = arctan(8.1 / 3.2)
θ ≈ 68.71 degrees
Therefore, the direction of the puck's velocity at t = 0.50 s relative to the +x axis is approximately 68.71 degrees.