Instructions:
PUKE in your note book the following problems.
An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?
A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?
If you want to see it
http://www.youtube.com/watch?v=s4tuvOer_GI&sns=tw via @youtube
Will this be due tomorrow in class?
ReplyDeleteI forgot my notebook in my locker, would it be alright if i did it on a separate sheet and copied it into tomorrow?
ReplyDeleteAbsolutely.
ReplyDeleteThis is not due tomorrow also please if you are posting to the adopt a physicist only post one question at a time. Thanks.
ReplyDeleteIf we posted all of our questions to adopt a physicist at once, should we redo them?
ReplyDelete-Katie M.
I share this question...
Delete-Robyn Kimmel
If you already posted them, I would assume that he already looked at them and that it's fine for those since it wasn't specified until now. For future questions that you post, though, he definitely wants them one at a time.
DeleteNo thats ok. Sorry for the delat
Deletewhen is this due ?
ReplyDeleteHe said that they're not due tomorrow, and by the looks of his last tweet, it could be possible that he won't be here tomorrow either. That gives us the class time in the cafeteria to PUKES through the problems with our classmates if we get stuck.
DeleteDoes that mean he is collecting it as soon as he gets back?
Delete-Skylar Young
Well, most of us wrote it in our notebooks, and we'll probably end up just discussing it in class. If he wants to collect it from a grade, he'll let us know.
Deletecool video. also, i was think about bringing my gopro camera on the rutgers trip, do you think that would be a good idea?
ReplyDelete-Will F
YEs it should be cool. At least that way I can see some of it. We can post it online for those students thatwere not able to go.
DeleteFor the first problem, is the starting velocity 0 or does it start at 3.2 or is it something completely different?
ReplyDeleteI also had this question, but I think it may be zero
Delete-Kyle O
The starting velocity is zero.
DeleteIn this problem is the acceleration 3.2 and not 9.8?
Delete-Skylar Young
Skylar the acceleration is 3.2 m/s2 for the first problem. I had put 9.8 at first but then looking back at the problem it stated that it was 3.2m/s2
DeleteWhat does uniformly mean?
ReplyDeleteNo jerk. The acceleration is constant.
DeleteFor the second one, I'm unsure about what equation to use? Am I supposed to find velocity? Cause you can't find acceleration without velocity and you can't find elicits without acceleration.
ReplyDeleteOk. You are looking for a. But if you write the equation a=change in v over t. You know the starting velocity-0. But you do t know the final v . So can you use another equation to find v f? Or draw out the picture that should help.
DeleteI'm still confused on number 2 I drew a picture but am still unsure of how to find an equation that will give you acceleration that doesn't have velocity.
ReplyDeleteYou could easily find the velocity first, and use that to find the acceleration. Think about it. Velocity = Change in Position / Change in Time. Plug and Chug. You now have the velocity, and you know that acceleration = Change in Velocity / Change in Time. Plug and Chug and you have your solution.
DeleteThis helped me so much. Thanks for saying that the change in position over the change in time is velocity. I was only using my other velocity formula.
DeleteBe careful what velocity have you solved. Is that the AVG? if you use Change in position over time then yes its the average. Vavg=(Vo+V)/2
DeleteWait if we have to find the final velocity what equation would we have to use? Because the velocity=change in p/ change in t is average. Can we move velocity around so it can be 2(Vavg)-Vo=Vf
DeleteFor the third question, instead of just finding the Vf then doing P+C, wouldn't we be finding the Po or P instead? Then apply that to find the distance.
ReplyDeleteWhich question are you referring to? The third one is only asking for the final velocity.
DeleteNumber 3 is asking for the final velocity and the distance he travelled. We know time, acceleration, V nought (you can call it zero), and P nought (also zero just to make the math easier. From here, you can find the final velocity with
DeleteV=a*t+(V nought)
To find the distance he travelled, just use
P=1/2at squared (V nought and P nought, being zero, would just cancel out).
-Kevin Meglathery
in the video, what is the use of the checkered square on the side of his face? and what causes him to stop so quickly when he was testing the fastest speed a pilot could eject at? was it water?
ReplyDeleteIt was a combination of water and rail brakes. I think the waters primary purpose was to cool the breaks.
Deletemr crane, like dr. stapp, would you put your life on the line for physics?
ReplyDeleteAbsolutely! Get me a cinder block and a bed of nails.
ReplyDeleteHere are the answers to 1&2:Given:
Knowns
a = +3.2 m/s2
t = 32.8 s
vi = 0 m/s
Unknown:
d = ??
d = vi*t + 0.5*a*t2
d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2
d = 1720 m
Problem 2
Knowns:
d = 110 m
t = 5.21 s
vi = 0 m/s
Find:
a = ??
d = vi*t + 0.5*a*t2
110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2
110 m = (13.57 s2)*a
a = (110 m)/(13.57 s2)
a = 8.10 m/ s2
Is this going to be collected right away or are we going to go over it first? Because I am confused with a few things
ReplyDeleteMr. Crane, will you be at school tomorrow?
ReplyDeleteCan someone please explain how 8.10 was gotten for acceleration for number 2. My answer was 4.05 m/s2. Here is what I did.
ReplyDeleteI know time which is 5.21s, Vo= 0m/s and d=110m. I used the equation v= delta p / delta to find velocity final and I got 21.1 m/s. then used the equation a=delta v / delta a and got 4.05 m/s2
Like Mr. Crane posted, change in position over time gives you the AVERAGE velocity. The velocity is not uniform (there is uniform acceleration, which you are trying to find), which is why can't use that formula to find V final, and then find acceleration. However, V naught is 0, so the average is half of V final because you can find the average by adding the initial and final velocity and dividing by 2.
DeleteInstead of using that formula to find the acceleration (because if you did not have V naught=0 you would not be able to find V final directly from V average to use the formula change in velocity/change in time), you could use P=1/2*at^2+V naught*t+P naught and rearrange it to find the acceleration.
On another note, for number 2, why is there 3 sig figs in the answer because 110 m has only 2 sig figs?
-Billy Potts
Thank you this really helped a lot.
DeleteI was just wondering if we will be able to go over this work because i had a few questions about the feather one.
ReplyDelete- Skylar Young
For this one you were not given time. You have no equation that will give you what you need. But there is a way to solve it. As long as you have PUKE. You are good. Then we can discuss it in class tomorrow.
DeleteFor the fourth question are the numbers given the starting and ending velocity or position?
ReplyDelete-Skylar Young
The units for that question is m/s so you know that it has to be the starting and ending velocity.
DeleteYou have to find the starting and the ending velocity i believe
DeleteWhat are the units? What do the measure?
ReplyDeleteSuppose John Stapp, instead of given a large strap and things to hold him from flying out, he is only given a wingsuit, approximately how far would Mr. Stapp fly?
ReplyDeleteFor the picture should we label as much as possible, or just as much as we need?
ReplyDeleteYes. The more you label, the more you know, and the more you know, the easier it is to find the answer.
Delete-Kevin Meglathery
Sam always label everything you know but make sure to also label the unknows. This will help you in the the rest of the PUKE if you know what you need to find and what you already have.
ReplyDeleteI think I'm just overlooking this, but for number 5 to plug into the new formula, how are you supposed to find the v?
ReplyDeleteThe formula can be used to find V (final) in #5, and then you plug it into one of the other formulas to find time (Like t=Change in V/a). Since the object in #5 is dropped, V naught is 0.
Delete-Billy Potts
I found the new equation so much eaiser to find the answer to #5 so glad you showed us it!
ReplyDelete