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course Phy 201
Problems:Show how you could use the information that a mile is 5280 feet, an inch is 2.54 cm, an hour is 60 minutes and a minute is 60 seconds, in order to determine how many miles / hour corresponds to a meter / second.
-feet into inches, inches into cm, cm into mm.
Right, but you need to show the details. See the solutions below.
I took the Interstate to work today. I slowed from about 18 meters / second to a stop in about 12 seconds. How fast was I moving, on the average? Can you give a single quantity that describes how quickly my velocity was changing?
Average rate=change in A/Change in B
Average rate=18m/12sec=1.5m/s good
-Average Velocity=Average rate of change of velocity with respect to clock time
1.5m/s good, but this is Average rate of change of velocity with respect to clock time (as you say), not average velocity (as you also appear to say). Be sure you have the terminology correct.
Before making the turn into VHCC, it took about 3 seconds to slow from 35 mph to 30 mph, another 3 seconds from 30 mph to 25 mph and another 3 seconds from 25 mph to 20 mph. What is a single quantity that describes how quickly my velocity was changing?
-5mph/3sec=1.66mph/sec good
A graph of rubber band tension vs. length, for a certain rubber band, isn't really a straight line. However between lengths 7 cm and 10 cm the graph is pretty well approximated by a straight line between the points (7 cm, 0 Newtons) and (10 cm, 4 Newtons).
* If that rubber band is stretched out between the points (4 cm, 9 cm) and (10 cm, 4 cm), what will be its length, and what will be its tension?
6cm, 5cm
Those are the legs of the right triangle whose hypotenuse is the length of the rubber band. But you have to find the hypotenuse.
* What are the x and y components of the displacement vector r from the first point to the second?
6cm, 5cm
The second would be -5 cm.
* If the x and y components are both divided by the length of the vector, what are the results? What is the sum of the squares of the results?
* What are the components of the unit vector u in the direction of the displacement vector r? (University Physics students should answer or attempt to answer; General College Physics students are invited but not yet required to answer this question)
* What vector do you get if you multiply u by the tension? (University Physics students should answer or attempt to answer; General College Physics students are invited but not yet required to answer this question)
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Compare your answers with the given solutions, and let me know if you have questions:
Answers to problems from notes 100825
Show how you could use the information that a mile is 5280 feet, an inch is 2.54 cm, an hour is 60 minutes and a minute is 60 seconds, in order to determine how many miles / hour corresponds to a meter / second.
There are 5280 ft in a mile and 12 inches in a foot. Therefore
1 mile = 5280 ft = 5280 ft * 12 in / ft = 63 360 in
There are 2.54 cm in an inch so
1 mile = 63 360 in * 2.54 cm / in = 160 933.4 cm
There are 100 cm in a meter so
1 mile = 160 933.4 cm * 1 meter / (100 cm) = 1609.334 m.
This entire series of calculations can be summarized in one calculation:
1 mile * 5280 ft / mile * 12 in / ft * 2.54 cm / in * 1 m / (100 cm) = 1609.334 m.
Now to calculate how many miles / hour correspond to a meter / second:
There are 3600 seconds in a hour so
1 m / s = 1 m * (1 mile / 1609.334 m) / s * 3600 s / h) = 3600 miles / (1609.334 h) = 2.24 miles / h, to three significant figures.
The entire conversion of a meter / second to miles / hour can be summarized in one calculation as
1 m / s * 100 cm / m * 1 inch / (2.54 cm) * 1 foot / (12 inches) * 1 mile / 5280 ft * 3600 s / hr = 2.24 miles / h, approximated to three significant figures.
I took the Interstate to work today. I slowed from about 18 meters / second to a stop in about 12 seconds. How fast was I moving, on the average? Can you give a single quantity that describes how quickly my speed was changing?
My speed changed from 18 m/s to 0 m/s, so averaging these two speeds would give a reasonable estimate of my average speed. We would get (18 m/s + 0 m/s) / 2 = 9 m/s.
My speed changed from 18 m/s to 0 m/s, a change of - 18 m/s, in 12 seconds. So my speed changed at a rate of -18 m/s / (12 s) = -1.5 (m/s) / 2.
Before making the turn into VHCC, it took about 3 seconds to slow from 35 mph to 30 mph, another 3 seconds from 30 mph to 25 mph and another 3 seconds from 25 mph to 20 mph. What is a single quantity that describes how quickly my velocity was changing?
Every three seconds my velocity decreased by 5 mph. So it was decreasing at a rate of (5 mph) / 3 s = 1.67 mph / sec. This tells us how quickly my velocity was changing. The answer to the question 'how quickly' would not indicate whether velocity was increasing or decreasing, but would be a positive number.
This is similar to a rate of change, except that a rate of change can be positive or negative.
My velocity was decreasing, so rate of change of velocity with respect to clock time would be -1.67 mph / sec.
A graph of rubber band tension vs. length, for a certain rubber band, isn't really a straight line. However between lengths 7 cm and 10 cm the graph is pretty well approximated by a straight line between the points (7 cm, 0 Newtons) and (10 cm, 4 Newtons).
If that rubber band is stretched out between the points (4 cm, 11 cm) and (10 cm, 4 cm), what will be its length, and what will be its tension? NOTE: The y coordinate of the first point is different than that originally given in the assignment. It was 9 cm, and is now 11 cm. If you solved using the 9 cm length your results will be different, but will follow the same sequence of steps as the given solution.
Its length would be the hypotenuse of a right triangle with legs | 10 cm - 4 cm | = 6 cm, and | 4 cm - 11 cm | = 7 cm.
By the Pythagorean Theorem the length would therefore be sqrt( (6 cm)^2 + (7 cm)^2 ) = sqrt( 85 cm^2) = 9.2 cm, approximately.
If you construct the graph of force vs. length you will find that when length is 9.2 cm, the force is about 2.9 Newtons.
What are the x and y components of the displacement vector r from the first point to the second?
The displacement vector has x component (10 cm - 4 cm) = 6 cm and y component (4 cm - 11 cm) = -7 cm.
If the x and y components are both divided by the length of the vector, what are the results? What is the sum of the squares of the results?
The length of the vector is about 8.1 cm, so x comp / length = 6 cm / (9.2 cm) = .65.
y comp / length = -7 cm / (9.2 cm) = -.75.
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