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 = +7.2 m/s and ax = +4.3 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +8.9 m/s and ay = -5.4 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(t)=vxi+ax*t
=7.2 + 4.3*.5
do the same for vy(t)
then theta=arctan(vy/vx)
magnitude v= sqrt(vx^2 + vy^2)
I figured it out, but thank you!
To find the magnitude v and direction θ of the puck's velocity at a time of t = 0.50 s, we can use the equations of motion.
The magnitude of velocity, v, can be found using the formula:
v = √(vx^2 + vy^2)
where vx and vy are the x and y components of the velocity respectively.
From the given information, we have vx = v0x + (ax)(t) and vy = v0y + (ay)(t).
Substituting the values,
vx = 7.2 m/s + (4.3 m/s^2)(0.50 s)
vx = 7.2 m/s + 2.15 m/s
vx = 9.35 m/s
vy = 8.9 m/s + (-5.4 m/s^2)(0.50 s)
vy = 8.9 m/s - 2.7 m/s
vy = 6.2 m/s
Now we substitute these values back into the magnitude equation:
v = √( vx^2 + vy^2)
v = √( 9.35 m/s)^2 + (6.2 m/s)^2
v = √( 87.2225 m^2/s^2 + 38.44 m^2/s^2)
v = √( 125.6625 m^2/s^2 )
v ≈ 11.2 m/s
So the magnitude of the velocity at t = 0.50 s is approximately 11.2 m/s.
Now, let's find the direction θ of the velocity relative to the +x axis.
θ = arctan(vy/vx)
θ = arctan(6.2 m/s/9.35 m/s)
θ = arctan(0.6631)
θ ≈ 33.8 degrees
Therefore, the direction θ of the puck's velocity at t = 0.50 s is approximately 33.8 degrees relative to the +x axis.