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 = +2.0 m/s and ay = -1.6 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
I've figured out that A.) is 1.732 m/s I just do not know how to figure out part B.) Help please
I figured it out!
To find the magnitude and direction of the puck's velocity at t = 0.50 s, we can use the equations of motion.
(a) Magnitude of the velocity (v):
The magnitude of the velocity can be calculated using the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
Given:
vx = v0x + ax * t, where v0x is the initial x-component of velocity, ax is the x-component of acceleration, and t is the time.
vy = v0y + ay * t, where v0y is the initial y-component of velocity, ay is the y-component of acceleration, and t is the time.
Substituting the given values into the equations:
vx = 0.8 m/s + 0.9 m/s^2 * 0.50 s = 1.3 m/s
vy = 2.0 m/s + (-1.6 m/s^2) * 0.50 s = 1.2 m/s
Using the equations above, we can calculate the magnitude of velocity:
v = sqrt((1.3 m/s)^2 + (1.2 m/s)^2)
v = 1.732 m/s (rounded to three decimal places)
Therefore, the magnitude of the velocity at t = 0.50 s is 1.732 m/s.
(b) Direction of the velocity (θ):
To find the direction of the velocity relative to the +x axis, we can use trigonometry.
θ = tan^(-1)(vy/vx)
Substituting the given values:
θ = tan^(-1)(1.2 m/s / 1.3 m/s)
θ = tan^(-1)(0.9231)
Using a calculator inverse tangent function:
θ = 43.64 degrees (rounded to two decimal places)
Therefore, the direction of the velocity at t = 0.50 s, relative to the +x axis, is approximately 43.64 degrees.