The figure below gives the path of a squirrel moving about on level ground, from point A (at time t = 0), to points B (at t = 4.00 min), C (at t = 8.00 min), and finally D (at t = 12.0 min). Both axes are marked in increments of 5.00 m (therefore the diagram is not drawn to scale). Consider the average velocities of the squirrel from point A to each of the other three points.
(a) Of the three average velocities, which has the least magnitude?
What is this average velocity in magnitude-angle notation?
m/s, ° (counterclockwise from the positive x axis)
(b) Which has the greatest magnitude?
What is this average velocity in magnitude-angle notation?
m/s, ° (counterclockwise from the positive x axis)
Nop
nop clop
To determine the magnitude and direction of average velocities, we need to calculate the displacements of the squirrel between each point and the time interval. Let's break it down step by step:
(a) To find the average velocity from point A to each of the other three points, we need to calculate the displacement and divide it by the time interval.
1. Firstly, we find the displacement between point A and point B. From the diagram, we can see that the squirrel moves horizontally along the positive x-axis and vertically along the positive y-axis. The displacement in the x-axis is 5 units to the right, and the displacement in the y-axis is 5 units upward. Therefore, the displacement between A and B is (5, 5), where the first number represents the change in x and the second number represents the change in y.
2. Next, we calculate the time interval between A and B, which is 4.00 min - 0 min = 4.00 min.
To calculate the average velocity, we divide the displacement by the time interval:
Average velocity AB = (5, 5) / 4.00 = (1.25, 1.25) m/min
3. Repeat the same steps to find the average velocities from point A to points C and D.
Average velocity AC = (15, 5) / 8.00 = (1.875, 0.625) m/min
Average velocity AD = (20, 10) / 12.0 = (1.667, 0.833) m/min
Now, we need to compare the magnitudes of these average velocities to determine which one has the least magnitude.
The magnitude of a vector (x, y) is given by the formula:
Magnitude = √(x^2 + y^2)
Calculate the magnitudes of average velocities AB, AC, and AD:
Magnitude AB = √[(1.25)^2 + (1.25)^2]
Magnitude AC = √[(1.875)^2 + (0.625)^2]
Magnitude AD = √[(1.667)^2 + (0.833)^2]
Compare the magnitudes to find the one with the least magnitude.
(b) Similarly, compare the magnitudes of AB, AC, and AD to find the one with the greatest magnitude.
Once you have determined the average velocity with the least and greatest magnitudes, we can convert them to magnitude-angle notation by using the inverse tangent function.
Magnitude-angle notation is given as (magnitude, angle), where the angle is measured counterclockwise from the positive x-axis.
To convert the average velocity vector into magnitude-angle notation:
For example, let's convert the average velocity AB from part (a). We have already calculated the magnitude, and now we need to find the angle.
Angle AB = arctan(y / x) = arctan(1.25 / 1.25)
Use the arctan function to find the angle in degrees.
Repeat this process for the other average velocities to convert them into magnitude-angle notation.