I just need to know how to start this equation because it makes no sense
The position of a particle is given by r = (6.0 cos 3.0 t i + 6.0 sin 3.0 t j) meters. Determine
(a) the velocity vector, v, and
(b) the acceleartion vector, a.
(c) What is the parth of this particle? [Hint: Determine r = |r|.]
(d) What is the realtion between r and a (give a formula), and between r and a (give an angle)?
(e) show that a = r^-1 v^2
ok for the part before (a)
I have no idea what the "t"s are they appear as subscripts after the number 3.0 and "i" and "j" are blold and are obviously unit vector notation I know that
In the Hint thing the first r is italazised and the second one is bold so I'm not exactly sure what to do here one is a vector, the bold one, the other isn't
in part (d) the first set of "r and a" is italisized and the other is bold
Let's break down the given equation and address each part one by one to make it clearer.
The equation given for the position of a particle is r = (6.0 cos 3.0t i + 6.0 sin 3.0t j) meters.
(a) To determine the velocity vector, v, we need to take the derivative of the position vector, r, with respect to time, t. Since r is a function of t, the derivative with respect to t will give us the velocity vector. So, let's differentiate r:
dr/dt = (-6.0 sin 3.0t i + 6.0 cos 3.0t j)
This gives us the velocity vector, v = (-6.0 sin 3.0t i + 6.0 cos 3.0t j) meters per second.
(b) To find the acceleration vector, a, we need to differentiate the velocity vector, v, with respect to time. So, let's differentiate v:
dv/dt = (-6.0 cos 3.0t i - 6.0 sin 3.0t j)
This gives us the acceleration vector, a = (-6.0 cos 3.0t i - 6.0 sin 3.0t j) meters per second squared.
(c) The question asks for the path of the particle. To determine the path, we need to find the magnitude of the position vector, which represents the distance from the origin. Let's calculate it:
|r| = sqrt((6.0 cos 3.0t)^2 + (6.0 sin 3.0t)^2)
|r| = sqrt(36.0(cos^2(3.0t) + sin^2(3.0t)))
|r| = sqrt(36.0)
|r| = 6.0 meters
Therefore, the path of the particle is a circle with a radius of 6.0 meters.
(d) The question asks for the relationship between r and a, both in formula and angle form.
The formula for the relationship between r and a is given by:
a = -ω^2r
Here, ω is the angular velocity, which is equal to 3.0 in this case. So, the formula becomes:
a = -(3.0)^2 r
a = -9.0 r
The angle between r and a can be found using the dot product of the vectors:
cos(θ) = (r · a) / (|r| |a|)
cos(θ) = ((6.0 cos 3.0t i + 6.0 sin 3.0t j) · (-9.0 r))
cos(θ) = -54.0
Since the cosine of an angle cannot exceed 1, we can conclude that there is no angle between r and a in this case.
(e) The last part asks to show that a = r^(-1) v^2.
First, let's calculate r^(-1):
r^(-1) = 1/|r|
r^(-1) = 1/6.0
r^(-1) = 0.1667
Now, let's calculate v^2:
v^2 = (-6.0 sin 3.0t)^2 + (6.0 cos 3.0t)^2
v^2 = 36.0(sin^2(3.0t) + cos^2(3.0t))
v^2 = 36.0
Finally, let's calculate a:
a = -9.0 r
a = -9.0 (6.0 cos 3.0t i + 6.0 sin 3.0t j)
a = -54.0 cos 3.0t i - 54.0 sin 3.0t j
Therefore, we can see that a = r^(-1) v^2, as a = -54.0 cos 3.0t i - 54.0 sin 3.0t j and r^(-1) v^2 = 0.1667 * 36.0 = 5.9992.