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 = +3.5 m/s and ax = +5.8 m/s2. The y components of the puck's initial velocity and acceleration are v0y = +2.8 m/s and ay = -2.0 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.
(a) Well, let's calculate the magnitude of the velocity first. We can use the Pythagorean theorem, which states that the square of the magnitude of the velocity is equal to the sum of the squares of its components. So, we have:
v^2 = v0x^2 + v0y^2
Simply plug in the given values:
v^2 = (3.5 m/s)^2 + (2.8 m/s)^2
Now do some math and calculate that:
v^2 = 15.25 m^2/s^2 + 7.84 m^2/s^2
Add those numbers together:
v^2 = 23.09 m^2/s^2
Finally, take the square root of both sides to find v:
v = √(23.09 m^2/s^2)
(b) Now let's find the direction of the velocity relative to the +x axis. To do that, we can use some trigonometry. We'll use the tangent function to find the angle è.
tan(è) = v0y / v0x
Plugging in the numbers:
tan(è) = 2.8 m/s / 3.5 m/s
Now find the angle by taking the inverse tangent (arctan) of that ratio:
è = arctan(2.8 m/s / 3.5 m/s)
Pop out your calculator and make that calculation, and you'll get the value of è in radians.
So, the magnitude of the velocity (v) is the square root of 23.09 m^2/s^2, and the direction (è) is the angle you found.
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.
(a) Magnitude v:
At time t = 0.50 s, we need to find the magnitude of the puck's velocity. The magnitude of the velocity vector is given by the equation:
v = sqrt((vx)^2 + (vy)^2)
Given:
vx = v0x + ax * t
vy = v0y + ay * t
v0x = +3.5 m/s
ax = +5.8 m/s^2
v0y = +2.8 m/s
ay = -2.0 m/s^2
t = 0.50 s
Substituting the given values into the equations:
vx = 3.5 m/s + 5.8 m/s^2 * 0.50 s = 3.5 m/s + 2.9 m/s = 6.4 m/s
vy = 2.8 m/s + (-2.0 m/s^2) * 0.50 s = 2.8 m/s - 1.0 m/s = 1.8 m/s
v = sqrt((6.4 m/s)^2 + (1.8 m/s)^2)
v = sqrt(40.96 m^2/s^2 + 3.24 m^2/s^2)
v = sqrt(44.20 m^2/s^2)
v ≈ 6.65 m/s
Therefore, the magnitude of the puck's velocity at t = 0.50 s is approximately 6.65 m/s.
(b) Direction θ:
To find the direction of the velocity, we can use the arctan function:
θ = arctan(vy / vx)
Substituting the known values:
θ = arctan(1.8 m/s / 6.4 m/s)
θ = arctan(0.28125)
Using a calculator, we find:
θ ≈ 15.42 degrees
Therefore, the direction of the puck's velocity at t = 0.50 s is approximately 15.42 degrees relative to the +x axis.
To find the magnitude v of the puck's velocity at time t = 0.50 s, we first need to determine the x and y components of the velocity at that time.
(a) Magnitude v of the velocity:
The magnitude v of the velocity is given by the formula:
v = sqrt(vx^2 + vy^2)
where vx is the x component of the velocity and vy is the y component of the velocity.
To find the x component of the velocity at t = 0.50 s, we can use the formula:
vx = v0x + ax * t
Substituting the given values:
vx = 3.5 m/s + (5.8 m/s^2) * 0.5 s
= 3.5 m/s + 2.9 m/s
= 6.4 m/s
To find the y component of the velocity at t = 0.50 s, we can use the formula:
vy = v0y + ay * t
Substituting the given values:
vy = 2.8 m/s + (-2.0 m/s^2) * 0.5 s
= 2.8 m/s + (-1.0 m/s)
= 1.8 m/s
Now, substituting the values of vx and vy into the formula for v:
v = sqrt((6.4 m/s)^2 + (1.8 m/s)^2)
= sqrt(40.96 m^2/s^2 + 3.24 m^2/s^2)
= sqrt(44.2 m^2/s^2)
≈ 6.65 m/s
Therefore, the magnitude v of the puck's velocity at t = 0.50 s is approximately 6.65 m/s.
(b) Direction è of the velocity:
To find the direction è of the velocity, we can use the formula:
θ = atan(vy/vx)
where θ is the angle between the velocity vector and the +x axis.
Substituting the values of vy and vx:
θ = atan((1.8 m/s)/(6.4 m/s))
= atan(0.28125)
Using a calculator or trigonometric table, we find that θ ≈ 15.3 degrees.
Therefore, the direction è of the puck's velocity relative to the +x axis at t = 0.50 s is approximately 15.3 degrees.
at t = 0
Vx = 3.5 and Ax = 5.8
so at t = .5
Vx = 3.5 + 5.8*.5 = 6.4
Vy = 2.8 and Ay = -2
so at t = .5
Vy = 2.8 -2*.5 = 1.8
|v| = sqt(6.4^2 + 1.8^2)
= 6.65 m/s
tan theta = 1.8/6.4
so theta = 15.7 degrees north of east