Find the following for path D in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.
Solution:
Part A
Looking at path D, we can see that it moves 2 blocks downward, 6 blocks to the right, 4 blocks upward, and 1 block to the left. Thus, the total distance of path D is
Looking at the initial and final position of path D, the final position is 5 blocks to the right or 600 meters to the right of the initial position, and 2 blocks or 240 meters upward from the initial position. Refer to the figure below.
Using the right triangle, we can solve for the displacement using the Pythagorean Theorem.
Find the following for path C in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.
Solution:
Part A
Looking at path C, it moves 1 block upward, 5 blocks to the right, 2 blocks downward, 1 block to the left, 1 block upward, and 3 blocks to the left. So, the total distance is
It can be seen from the figure that the end of path C is just one block to the right from the starting point. Therefore, the magnitude of the displacement is
(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R’=A−B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R”=B−A=−R’ ). Show that this is the case.
Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)
Solution:
Consider Figure 3-6A below with B drawn first before A.
Compute for the value of angle β by adding 20° and the complement of 40°. This is by simple geometry.
(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t=2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.
Solution:
Part A
To find the slope at t=2.5 s, we need the position values at t= 0 s and t=5 s. When t = 0 \ \text{s}, x = 0 \ \text{m}, and when t = 5 \ \text{s}, x = 17.5 \ \text{m}. The velocity at t=2.5 s is
Construct the position graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.
(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) The velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.
Note that the values are approximated to satisfy the given acceleration in the problem statement. The values may differ from one’s answer due to some uncertainties of a graph.
Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s.
Solution:
We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.
To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.
Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertainties when using graphs. The difference is not really significant for this case.
Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s. Assume all values are known to 2 significant figures.
Solution:
We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.
To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.
Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertain when using graphs. The difference is not really significant for this case.
(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t=20 s. (b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s2 .
Solution:
Part A
Figure A shows the approximate slope of the curve at time 20 seconds.
To solve for the slope of this line, we need to approximate by using two points. In this case, we shall use the points at time 15 seconds and 25 seconds.
Approximately, when t=15\ \text{s}, the position is x=1000\ \text{m}, and when t=25\ \text{s}, the position is x=2150\ \text{m}. Thefore,
One can immediately figure out from the given graph that it is a straight line. The slope of the line can be computed by using any two points in the line.
Here, v=15\ \text{m/s} when t=0\ \text{s}, and v=40 \ \text{m/s} when t=5\ \text{s}. The acceleration is
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