#$&*
course Phy 241
8/31 1 This is the assignment I recieved last night. I am sort of confused, is this the only assignment I am behind(besides questions turned in each class)on?????? Please let me know what else is expected of me thus far in the class. Thanks Again
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.
Answer
mile = 5280ft
inch = 2.54 cm
1 hr = 60 min
1 min = 60 sec
We can then use this info and common sense to convert,
A) 5280ft(1 mile)* 12(# inches in ft) = 63,360(# inches in mile)
63,360 (# inches in 1 mile)* 2.54(# cm per inch) = 160,934.4(# cm in mile)
160,934.4(# cm in mile)/100(# cm in a meter) = 1,609.344 !!!meters per mile!!!!
B)1 hr = 60 min
60(# of min in hr)*60(# of sec per min) = 3600 !!!# of sec per hr!!!
So that 1,609.344(meters per mile )/3600(sec per hr)= .44704
Knowing this if you have measurement in miles per hr it is easy to convert.
Ex) 60 mph = 60 miles/1 hr = 60/1 * 1,609.344(meters per mile )/3600(sec per hr) =
96,560.64(# meters in 60 miles)/3600(# sec in 1 hr)=26.8224 meters per sec
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?
Answer
If we look at a graph to help visualize our problem we would see x axis labeled “t in sec”, y axis labeled velocity (meters/sec). Our starting point is found at (t, 18) and ending point is (t+12, 0). Even though very unlikely I am going under the impression that out fn in constant fn, decreasing at a constant non changing rate. If this is so and we add another point at (t+6, ?) we can find the y value (because we are assuming the rate decreases constantly from t to t+12) by finding slope of fn.
Slope = rise/run = 0 – 18/12 = -1.5
So now to find y value at x = t+6 we can rearrange the same formula for slope ….
rise = slope*run = -1.5*((t+6)-t)(which is `dt or change in t) = -9
!!!We must remember this is not the y value, but only the change in y. Since y or fn value at t is 18, we add y and `dy. which is 18+(-9) = 9, so now y value is 9 meters per sec !!!!
So the ave speed is 9 m/s. Even if we use formula
average velocity = final displacement/total time taken, we have
ave velocity = 108m/12 = 9 m/s
?????So which is the correct way for us to approach this question, ave speed or ave velocity, also I know graph was not needed that is just a method I use to help see problem????????
Your approach was very sophisticated. The situation can be reasoned out more simply, but there's nothing wrong with using a powerful method. Well done.
The linear assumption is fairly valid for braking. It's not difficult to achieve pretty much constant acceleration during braking, as opposed to speeding up (where it is very unnatural and difficult to achieve).
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?
Answer
Acceleration is the change in velocity, (`dv), divided by the change in time, (`dt). So, a = Δv/Δt
Plugging in our values into our formula we get ,a=(v-vo)/t
= (20 – 35)/9 = -15/9 = -1.6667 mph/s????? I am not sure I am doing this correctly????
That's it.
If this correct then to find just mph would it look like this….
=(20 – 35)/(9/3600) = -15/(9/3600) = -6000 mph
Use units throughout your calculation. If you do you will find that you get -6000 mph / h.
???? Again I am not sure this is correctly setup or if I am doing this totally correctly.Even though I feel like its all fairly easy to understand????
You're doing great.
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?
Answer
To find length we could use dist formula or simply use Plathgarium Therorm.
Plathgarium Therorm:
6^2 + 5^2 = c^2
c^2 = 36 + 25
c = sqrt(61) or approx. 7.8102
Looking at graph of length in cm(x axis) vs Newtons(Force), it seems that force needed is approx. 1.25-1.5 Newtons.!!!!This is not exact, only an approx.!!!!!!!
* What are the x and y components of the displacement vector r from the first point to the second?
Answer
6 cm is vector measurement in x direction, 5 cm is vector measurement in y direction.
* 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?
Answer
<6cm, 5cm>/7.8 = <.7692, .6410>
* 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)
Answer
?????I am not sure what question is asking?????
* 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)
Answer
<.7692, .6410>*1.25(approx. of Newtons needed) = <.9615, .80125>
That's it. Excellent work.