At one instant a bicyclist is 58.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 35.0 s later, the cyclist is 58.0 m due north of the flagpole, going due east with a speed of 13.0 m/s. For the cyclist in this 35.0 s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration? (Give all directions as positive angles relative to due east, where positive is measured going counterclockwise.)
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To solve this problem, we will break it down into several steps:
Step 1: Find the displacement of the cyclist.
To find the displacement, we need to determine the change in position of the cyclist from the starting point to the ending point. We can determine the displacement by subtracting the initial position from the final position.
The initial position of the cyclist is 58.0 m due east, and the final position is 58.0 m due north. Since we are given the positions in terms of cardinal directions (east/north), we need to convert them into vector form.
Let's set up our coordinate system:
- East is the positive x-direction
- North is the positive y-direction
The initial position vector is (58.0 m, 0 m), and the final position vector is (0 m, 58.0 m).
To find the displacement vector, we subtract the initial position vector from the final position vector:
Displacement = (0 m, 58.0 m) - (58.0 m, 0 m)
= (-58.0 m, 58.0 m)
Therefore, the displacement vector is (-58.0 m, 58.0 m).
Step 2: Find the magnitude and direction of the displacement.
To find the magnitude of the displacement, we can use the Pythagorean theorem:
Magnitude of the displacement = √((-58.0 m)^2 + (58.0 m)^2)
= √(3364.0 m^2 + 3364.0 m^2)
= √(6728.0 m^2)
≈ 82.0 m
To find the direction, we can use trigonometry. Since the displacement vector points in the northeast quadrant, we can use the inverse tangent function (tan⁻¹) to find the angle:
Direction of the displacement = tan⁻¹(58.0 m / 58.0 m)
≈ 45.0°
Therefore, the magnitude of the displacement is approximately 82.0 m, and the direction is approximately 45.0° relative to due east.
Step 3: Find the average velocity.
The average velocity is defined as the displacement divided by the time interval:
Average velocity = Displacement / Time interval
In this case, the time interval is given as 35.0 s.
Average velocity = (-58.0 m, 58.0 m) / 35.0 s
= (-58.0 m / 35.0 s, 58.0 m / 35.0 s)
= (-1.66 m/s, 1.66 m/s)
Therefore, the average velocity is approximately (-1.66 m/s, 1.66 m/s).
Step 4: Find the magnitude and direction of the average velocity.
To find the magnitude of the average velocity, we can use the Pythagorean theorem:
Magnitude of the average velocity = √((-1.66 m/s)^2 + (1.66 m/s)^2)
= √(2.7556 m^2/s^2 + 2.7556 m^2/s^2)
= √(5.5112 m^2/s^2)
≈ 2.35 m/s
To find the direction, we can use trigonometry. Since the average velocity vector points in the northeast quadrant, we can use the inverse tangent function (tan⁻¹) to find the angle:
Direction of the average velocity = tan⁻¹(1.66 m/s / 1.66 m/s)
≈ 45.0°
Therefore, the magnitude of the average velocity is approximately 2.35 m/s, and the direction is approximately 45.0° relative to due east.
Step 5: Find the average acceleration.
The average acceleration is defined as the change in velocity divided by the time interval:
Average acceleration = Change in velocity / Time interval
In this case, the change in velocity is 13.0 m/s - (-1.66 m/s) = 14.66 m/s, and the time interval is given as 35.0 s.
Average acceleration = (14.66 m/s) / (35.0 s)
≈ 0.419 m/s^2
Therefore, the average acceleration is approximately 0.419 m/s^2.
Step 6: Find the direction of the average acceleration.
Since the average acceleration is a scalar, it does not have a specific direction. It is just a measure of the rate of change of velocity.
Therefore, the magnitude of the average acceleration is approximately 0.419 m/s^2, and it does not have a specific direction.
In summary:
(a) The magnitude of the displacement is approximately 82.0 m.
(b) The direction of the displacement is approximately 45.0° relative to due east.
(c) The magnitude of the average velocity is approximately 2.35 m/s.
(d) The direction of the average velocity is approximately 45.0° relative to due east.
(e) The magnitude of the average acceleration is approximately 0.419 m/s^2.
(f) The average acceleration does not have a specific direction; it is only a scalar quantity.