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 = +6.6 m/s and ax = +2.0 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +9.5 m/s and ay = -6.5 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.
Jfjfroro
To find the magnitude v of the puck's velocity at time t = 0.50 s, we can use the following equation:
v = √(vx^2 + vy^2)
In this equation, vx is the x-component of the velocity and vy is the y-component of the velocity.
Given:
vx = v0x + ax * t
= 6.6 m/s + 2.0 m/s^2 * 0.50 s
= 7.6 m/s
vy = v0y + ay * t
= 9.5 m/s + (-6.5 m/s^2) * 0.50 s
= 6.25 m/s
Plugging these values into the equation, we have:
v = √(7.6 m/s)^2 + (6.25 m/s)^2
= √(57.76 m^2/s^2 + 39.0625 m^2/s^2)
= √(96.8225 m^2/s^2)
= 9.84 m/s
Therefore, the magnitude of the puck's velocity at time t = 0.50 s is 9.84 m/s.
To find the direction θ of the puck's velocity, we can use the following equation:
θ = arctan(vy/vx)
Given:
vx = 7.6 m/s
vy = 6.25 m/s
Plugging these values into the equation, we have:
θ = arctan(6.25 m/s / 7.6 m/s)
= arctan(0.82)
Using a calculator or trigonometric table, we find:
θ ≈ 39.19 degrees
Therefore, the direction of the puck's velocity at time t = 0.50 s, relative to the +x axis, is approximately 39.19 degrees.