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 = +8.3 m/s and ax = +8.8 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +9.6 m/s and ay = -6.1 m/s2. Find (a) the magnitude v and (b) the direction è of the puck's velocity at a time of t = 0.50 s. Specify the direction relative to the +x axis.
thanks guys for the help.
now my mom wont beat me since I've done my work
(my dad cant beat me because he went to go get milk a few years ago)
V = Vo + at.
V(hor) = 8.3 + 8.8*0.5 = 12.7m/s.
V(ver) = 9.6 + (-6.1)*0.5 = 6.55m/s.
a. V^2 = (12.7)^2 + (6.55)^2 = 204.19,
V = sqrt(204.19) = 14.3m/s = Magnitude.
b. tanA = Y/X = 6.55 / 12.7 = 0.516,
A = 27.3Deg. = Direction.
V = 14.3m/s @ 27.3Deg.
To find the magnitude and direction of the puck's velocity at t = 0.50 s, we can use the equations of motion.
Step 1: Find the x- and y-components of the velocity at t = 0.50 s.
Given:
v0x = +8.3 m/s (initial x-component velocity)
ax = +8.8 m/s^2 (x-component acceleration)
v0y = +9.6 m/s (initial y-component velocity)
ay = -6.1 m/s^2 (y-component acceleration)
t = 0.50 s (time)
Using the equation of motion: v = v0 + a*t
For the x-component:
vx = v0x + ax * t
= 8.3 m/s + 8.8 m/s^2 * 0.50 s
= 12.7 m/s(direction: along the positive x-axis)
For the y-component:
vy = v0y + ay * t
= 9.6 m/s + (-6.1 m/s^2) * 0.50 s
= 6.05 m/s (direction: along the positive y-axis)
Step 2: Find the magnitude and direction of the velocity at t = 0.50 s.
Using the Pythagorean theorem:
v^2 = vx^2 + vy^2
v = sqrt(vx^2 + vy^2)
= sqrt((12.7 m/s)^2 + (6.05 m/s)^2)
= sqrt(161.29 + 36.60)
= sqrt(197.89)
≈ 14.06 m/s
To find the direction θ (relative to the +x axis), we can use the inverse tangent function:
θ = atan(vy/vx)
= atan(6.05 m/s / 12.7 m/s)
≈ atan(0.4764)
≈ 26.18 degrees
Therefore,
(a) The magnitude of the velocity v ≈ 14.06 m/s.
(b) The direction of the velocity θ ≈ 26.18 degrees (relative to the +x axis).
To find the magnitude and direction of the puck's velocity at t = 0.50s, we can use the kinematic equations for motion.
(a) Calculate the x and y components of the velocity at t = 0.50s using the equations:
vx = v0x + ax * t and vy = v0y + ay * t
Substituting the given values:
vx = 8.3 m/s + 8.8 m/s^2 * 0.50 s = 8.3 m/s + 4.4 m/s = 12.7 m/s
vy = 9.6 m/s + (-6.1 m/s^2) * 0.50 s = 9.6 m/s - 3.05 m/s = 6.55 m/s
(b) To find the magnitude of the velocity (v), use the Pythagorean theorem formula:
v = sqrt(vx^2 + vy^2)
Substituting the values we calculated:
v = sqrt((12.7 m/s)^2 + (6.55 m/s)^2) = sqrt(161.29 m^2/s^2 + 42.90 m^2/s^2)
v = sqrt(204.19 m^2/s^2) = 14.29 m/s (rounded to two decimal places)
(c) To find the direction (theta), use the inverse tangent function:
theta = arctan(vy / vx)
Substituting the values we calculated:
theta = arctan(6.55 m/s / 12.7 m/s) = 28.71 degrees (rounded to two decimal places)
So, at t = 0.50s, the magnitude of the puck's velocity is 14.29 m/s and the direction relative to the +x axis is 28.71 degrees.