A particle travels counterclockwise around the origin at constant speed in a circular path that has a diameter of 2.50 m, going around twice per second. Point Q is on the path halfway between points P and R, which lie on the x axis and the y axis, respectively. State the following velocites as vectors in polar notation. (a) What is the particle's average velocity over the interval PR? (b) What is its average velocity over the interval PQ? (c) What is its instantaneous velocity at point P?
P coordinates: (1.25, 0)
R coordinates: (0, 1.25)
Q is directly between the two
For A, 11.1 and 45 degrees is incorrect
For B, 14.5 and 22.5 degrees is incorrect
For C, 15.7 is correct, but there is also an answer in degrees I need.
PLEASE HELP!
To find the velocities as vectors in polar notation, we need to consider the direction and magnitude of the velocities.
(a) Average velocity over the interval PR:
To find the average velocity over the interval PR, we need to calculate the change in position and divide it by the time taken. The change in position from P to R can be found by subtracting the coordinates of P from the coordinates of R:
Δx = (0 - 1.25) = -1.25
Δy = (1.25 - 0) = 1.25
The time taken to travel from P to R is given as 1 second, as the particle goes around the circular path twice per second.
Average velocity = (Δx / t)î + (Δy / t)ĵ
= (-1.25 / 1)î + (1.25 / 1)ĵ
= -1.25 î + 1.25 ĵ
To convert this average velocity to polar notation, we can use the magnitude and direction of the velocity. The magnitude of the velocity can be found using the Pythagorean theorem:
|V| = √((-1.25)^2 + (1.25)^2)
≈ √(1.5625 + 1.5625)
≈ √3.125
≈ 1.77 m/s
The direction of the velocity can be found using the tangent function:
θ = arctan(Δy / Δx)
= arctan(1.25 / -1.25)
= arctan(-1)
≈ -45°
Therefore, the average velocity over the interval PR can be represented as a vector in polar notation as 1.77 m/s at an angle of -45°.
(b) Average velocity over the interval PQ:
To find the average velocity over the interval PQ, we need to calculate the change in position and divide it by the time taken. The change in position from P to Q can be found by subtracting the coordinates of P from the coordinates of Q:
Δx = (1.25 - 0.625) = 0.625
Δy = (0.625 - 0) = 0.625
The time taken to travel from P to Q is given as 0.5 second, as the particle goes around the circular path twice per second.
Average velocity = (Δx / t)î + (Δy / t)ĵ
= (0.625 / 0.5)î + (0.625 / 0.5)ĵ
= 1.25 î + 1.25 ĵ
To convert this average velocity to polar notation, we can use the magnitude and direction of the velocity. The magnitude of the velocity remains the same as it is the same for both P and Q and can be calculated as:
|V| = √((1.25)^2 + (1.25)^2)
≈ √(1.5625 + 1.5625)
≈ √3.125
≈ 1.77 m/s
The direction of the velocity can be found using the tangent function:
θ = arctan(Δy / Δx)
= arctan(0.625 / 0.625)
= arctan(1)
≈ 45°
Therefore, the average velocity over the interval PQ can be represented as a vector in polar notation as 1.77 m/s at an angle of 45°.
(c) Instantaneous velocity at point P:
The instantaneous velocity at point P can be calculated by finding the tangent to the circular path at that point. Since the particle is traveling counterclockwise around the origin, the tangent at point P is perpendicular to the x-axis.
The magnitude of the instantaneous velocity also remains the same as it is the same for both P and Q and can be calculated as:
|V| = 1.77 m/s (same as above)
The direction of the velocity can be represented by the angle formed between the velocity vector and the positive x-axis. Since the tangent at point P is perpendicular to the x-axis, the angle is 90 degrees.
Therefore, the instantaneous velocity at point P can be represented as a vector in polar notation as 1.77 m/s at an angle of 90°.