Heyy guys. For number 3 on last years test I set up the equation like this.... d1+d2=5334 and substituted each d for the equation d=v0t +1/2at. and got to 1/2a(t1)+v0(t2)=5334...but now i don't know where to go with it..can anyone help?
lauren! okay, so your initial equatoin is 100% correct, now since you don't know vo your having trouble. What you do is, your FINAL VELOCITY of the (d1) will be EQUAL to the INITIAL VELOCITY of (d2). why? Remember that rocket problem? You don't just suddenly "switch," you need to slow down / accelerate and stuff.
Okay, so once you have that, you know that the average acceleration is the change in velocity OVER the change in time. In other words, a=v/t. Now, you move the "t" and you get v=at. Since you know the acceleration for d1, your "vo" will equal (12.2m/s2)(t1). T1 because that is the time for distance one, and the final velocity of distance one is the inital velocity of distance 2!
The reason: Why we had to get vo = to (12.2)(t1) was if we left it as vo, we would have THREE UNKNOWNS and only TWO EQUATIONS. Makeing it impossible for us to solve, by converting it to (12.2)(t1). We only have TWO UNKNOWNS (t1,t2) and TWO EQUATIONS! ( (1/2)at^2 + vot = 5334 AND t1+t2 = 90. )
Plug those two into your calculater using the 'and' which you can find under Catalog (hit catalog, and hit the key with the "A" printed above it) and then make sure when you finished typeing those you put the following:
EX: Solver(eq1 and eq2, {x,y}) And BAM! You will have time one and time two.
T1 : 5 seconds T2: 85 seconds.
(You can text me if you need more help btw, you know, since we're all bf n stuff :P)
If you are talking about 1d motion practice problems #2 sheet, then I think that the rocket on the way down is a faster time lapse because of the fact that even though it has basically the same distance to cover, it is not interrupted by any engine shutting off/turning on, thus having one continuous motion downward. Then, the only thing that can act upon the rocket is gravity.Also, it would take a longer time for the rocket to ascend into the air because velocity is going up while gravity(acceleration) is pointing downward. On the rocket's descent, velocity and gravity(acceleration) are both pointing down to the ground, thus having more of a pull.I believe that is the reason why it takes a shorter period of time to have the rocket return to the ground.
Ohh, okay Mihir, use the v2=vo2+2ad equation and since you know the height (-10) and the acceleration (-9.8) simply plug those in!
You'll get x^2 (since your SOLVING for final velocity) = 0^2 +2(-9.8(-10) and you get 14.
14 is the SAME as 2b, so the answer is the same.
Plus, remember that bullet problem Mr. Rhodino did on the board today? The INITIAL speed going UP is the SAME as the VELOCITY RIGHT BEFORE it hits the ground and stuff, only, it is negative (when considering vector), otherwise, the MAGNITUDE (number ONLY) is the same.!
hey can anyone explain to me how to start #4 on practice problems #2 sheet about the man and the attack dog? What formula would you use for this problem?
Auxillary equations: i need help with them. Can anyone explain how you know which variable to add time to? so if a red stone gets dropped and 2 seconds later, the black stone gets dropped, which one gets the time added to it and so on.
Anne, the first 2 asterisks are correct, but the last one, it should be the absolute value of velocity = speed, right? Cause velocity is just speed with a direction.
Ok when converting from distance vs. time to an velocity vs. time you find the slope right n then plot that slope for the amount of time, right? but what if the line passes through the x-axis, do you count that as one line with one slope? or do you have to plot a separate velocity for the part that is under the x-axis?
eunice park. i think what you are saying only applies when you find the area of each part. so like, velocity vs time --> distance vs. time. because even if the line goes past the x-axis, the slope is going to be the same.
QUESTIONS. sjaklfjasklfj #16 on the 1D motion practice MC worksheet. if one is thrown up and the other is thrown down, wouldn't the one thrown up have a different final velocity when it hits the ground ? because... it's in the air for a longer amount of time ?
Eunice, you would count that as one slope for the line even if it passes through the X-axis. We did it on the first worksheet we got or on one of the ones we did together im pretty sure. look back and you'll see it.
You throw a ball down at x velocity. You throw another ball up at x velocity.
Whatever x is, gravity is constant. Say it takes 5 seconds to decelerate from x to 0, then because gravity is constant, it will take 5 seconds for the object to go from 0 back to x. When it hits x, it will be back at it's starting place, which is the same speed which the first object was thrown at.
Does that make sense? They will not hit the ground at the same time but they will hit the ground at the same speed.
For # 16 correct me If I'm wrong, but wouldn't they have the same velocity when they hit the ground because the ball that's thrown up gets to velocity=0 and then the acceleration of -9.8 increase the velocity at the same rate it would increase the velocity of one just thrown down.
For #16, if you still don't get it. Gravity is constant. ALWAYS (-9.8). And plus, small objects reach terminal velocity rather quickly, so despite the fact one is thrown up, it will come back to the same level as where ball one was thrown, so its displacement at that point is 0, and now they both go the SAME negative distance with the SAME initial velocity and the SAME (-9.8) acceleration.
So yeah, as David explained, the speed would be equal.
Anne, i confirm what Tim said. And lauren, it tells you that the TOTAL time is 90 seconds. So its just t1+t2 = 90
Lauren, you are at the 12.2 acceleration for t1 until the engine shuts off, then you maintain constant "v" for t2. And after t2 is done, you have traveled 5335 m in 90 seconds. So t1 + t2 = 90.
okay thanks Arjun I get it now. But instead of using 2 equations I just plugged in t1=90-t2 so everywhere that t1 was supposed to be I put in 90-t2 =]]
For #3 on the review worksheet, doesn't the question mean that the dog is 21m behind and waits for 6 seconds? so that the distance between the man and the dog is 31.2m, isn't it? since 21m+(1.7m/s x 6s)=31.2m
no, the distance between the man and the dog is 21 meters no matter what. The distance the dog has to travel to a certain point, as compared with the man, is longer.
so therefore the equations would be the distance of the dog-21=distance of the man
taehwan, so let's say the single dotted line is 21 m and the double dotted line represents the distance the man covers. if you think about it, the entire distance (distance of the dog) minus the distance of the man is going to equal 21m.
31.2m is the distance between the man and the dog after 6 seconds, but i don't think that will help you solve the time the man gets bitten by the dog.
I agree with taewhan on this one. Because it says the distance between te man and the dog is 21 meters. Then he waits 6 seconds. The man didn't stop runnig when the dog was waiting. So shouldt you have to take into account the distance he ran while the dog wa waitig. This is a poorly worded question.
Annie, for part 1b on the pop quiz, it is t= 1-1.6ish seconds, and 5-6 seconds.
As for part c, you just find the area above/below the lines and add/subtract accordingly to find the total displacement and where it will end up at t=7 seconds. Sorry I'm too lazy to do it all out here haha
OK Maggie. So whenever you have a problem which involves something being thrown up and it coming back down and you need to find the total time, you need use the displacement of the object not for d. Though it traveled up 40 and down 10, it ended up only 30 meters away from the starting point. That's where the 30 came in. The book explains it too.
Heyy guys. For number 3 on last years test I set up the equation like this.... d1+d2=5334 and substituted each d for the equation d=v0t +1/2at. and got to 1/2a(t1)+v0(t2)=5334...but now i don't know where to go with it..can anyone help?
ReplyDeletelauren! okay, so your initial equatoin is 100% correct, now since you don't know vo your having trouble. What you do is, your FINAL VELOCITY of the (d1) will be EQUAL to the INITIAL VELOCITY of (d2). why? Remember that rocket problem? You don't just suddenly "switch," you need to slow down / accelerate and stuff.
ReplyDeleteOkay, so once you have that, you know that the average acceleration is the change in velocity OVER the change in time. In other words, a=v/t. Now, you move the "t" and you get v=at. Since you know the acceleration for d1, your "vo" will equal (12.2m/s2)(t1). T1 because that is the time for distance one, and the final velocity of distance one is the inital velocity of distance 2!
The reason: Why we had to get vo = to (12.2)(t1) was if we left it as vo, we would have THREE UNKNOWNS and only TWO EQUATIONS. Makeing it impossible for us to solve, by converting it to (12.2)(t1). We only have TWO UNKNOWNS (t1,t2) and TWO EQUATIONS! ( (1/2)at^2 + vot = 5334 AND t1+t2 = 90. )
Plug those two into your calculater using the 'and' which you can find under Catalog (hit catalog, and hit the key with the "A" printed above it) and then make sure when you finished typeing those you put the following:
EX: Solver(eq1 and eq2, {x,y}) And BAM! You will have time one and time two.
T1 : 5 seconds
T2: 85 seconds.
(You can text me if you need more help btw, you know, since we're all bf n stuff :P)
Hey can anyone explain 2c to me? I don't understand why it is the same. Wouldn't it be slower on the way down?
ReplyDelete2c for what worksheet
ReplyDeleteIf you are talking about 1d motion practice problems #2 sheet, then I think that the rocket on the way down is a faster time lapse because of the fact that even though it has basically the same distance to cover, it is not interrupted by any engine shutting off/turning on, thus having one continuous motion downward. Then, the only thing that can act upon the rocket is gravity.Also, it would take a longer time for the rocket to ascend into the air because velocity is going up while gravity(acceleration) is pointing downward. On the rocket's descent, velocity and gravity(acceleration) are both pointing down to the ground, thus having more of a pull.I believe that is the reason why it takes a shorter period of time to have the rocket return to the ground.
ReplyDeleteHope this helps Mihir!!!
Ohh, okay Mihir, use the v2=vo2+2ad equation and since you know the height (-10) and the acceleration (-9.8) simply plug those in!
ReplyDeleteYou'll get x^2 (since your SOLVING for final velocity) = 0^2 +2(-9.8(-10) and you get 14.
14 is the SAME as 2b, so the answer is the same.
Plus, remember that bullet problem Mr. Rhodino did on the board today? The INITIAL speed going UP is the SAME as the VELOCITY RIGHT BEFORE it hits the ground and stuff, only, it is negative (when considering vector), otherwise, the MAGNITUDE (number ONLY) is the same.!
hey can anyone explain to me how to start #4 on practice problems #2 sheet about the man and the attack dog? What formula would you use for this problem?
ReplyDeletesorry guys, i need help on #4 for the 1d motion review problems, not the practice problems #2 sheet. OOPS?!?
ReplyDeletei dont get why the distance of the dog - distance of the man would equal 21
ReplyDeleteAuxillary equations: i need help with them. Can anyone explain how you know which variable to add time to? so if a red stone gets dropped and 2 seconds later, the black stone gets dropped, which one gets the time added to it and so on.
ReplyDeleteNeha, when you are setting the equations equal to each other, ask yourself "Which stone is traveling for longer?"
ReplyDeleteIn your example, the red stone is traveling longer, so it is (t+2)^2, while the black stone is t^2.
I would just like to confirm some things:
ReplyDelete*d, a, & v can be positive/negative (right?)
*d is negative when an object is moving downward/backward (right?)
*the absolute value of acceleration = speed (right?)
for 2d on the practice test, would the velocity coming down have to be negative? i did that and got 2.86s for the total time is that right?
ReplyDeletedoes anyone know a quick way to do number 7 on the multiple choice without a calculator?
ReplyDeleteAnne, the first 2 asterisks are correct, but the last one, it should be the absolute value of velocity = speed, right? Cause velocity is just speed with a direction.
ReplyDeleteOk when converting from distance vs. time to an velocity vs. time you find the slope right n then plot that slope for the amount of time, right? but what if the line passes through the x-axis, do you count that as one line with one slope? or do you have to plot a separate velocity for the part that is under the x-axis?
ReplyDeleteThanks Tim =]
ReplyDeleteCan someone else confirm his answer?
Arjun- where does the second equation come from the one with the 90 in it??
ReplyDeleteHaha, sorry Anne. I mean the question to you, if you understood it... I'm quite confident that my answer is right.
ReplyDeleteeunice park.
ReplyDeletei think what you are saying only applies when you find the area of each part. so like, velocity vs time --> distance vs. time. because even if the line goes past the x-axis, the slope is going to be the same.
QUESTIONS. sjaklfjasklfj
#16 on the 1D motion practice MC worksheet. if one is thrown up and the other is thrown down, wouldn't the one thrown up have a different final velocity when it hits the ground ? because... it's in the air for a longer amount of time ?
Eunice, you would count that as one slope for the line even if it passes through the X-axis. We did it on the first worksheet we got or on one of the ones we did together im pretty sure. look back and you'll see it.
ReplyDeleteim also with annie on #16 its really confusing i have no idea why its the same.
ReplyDelete#16:
ReplyDeleteYou throw a ball down at x velocity. You throw another ball up at x velocity.
Whatever x is, gravity is constant. Say it takes 5 seconds to decelerate from x to 0, then because gravity is constant, it will take 5 seconds for the object to go from 0 back to x. When it hits x, it will be back at it's starting place, which is the same speed which the first object was thrown at.
Does that make sense? They will not hit the ground at the same time but they will hit the ground at the same speed.
thank you john!
ReplyDeleteannie i'm not talking about the area :P im talking about the slope silly.
For # 16 correct me If I'm wrong, but wouldn't they have the same velocity when they hit the ground because the ball that's thrown up gets to velocity=0 and then the acceleration of -9.8 increase the velocity at the same rate it would increase the velocity of one just thrown down.
ReplyDeleteDavid, thank you that helped me out.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteFor #16, if you still don't get it. Gravity is constant. ALWAYS (-9.8). And plus, small objects reach terminal velocity rather quickly, so despite the fact one is thrown up, it will come back to the same level as where ball one was thrown, so its displacement at that point is 0, and now they both go the SAME negative distance with the SAME initial velocity and the SAME (-9.8) acceleration.
ReplyDeleteSo yeah, as David explained, the speed would be equal.
Anne, i confirm what Tim said. And lauren, it tells you that the TOTAL time is 90 seconds. So its just t1+t2 = 90
ReplyDeleteLauren, you are at the 12.2 acceleration for t1 until the engine shuts off, then you maintain constant "v" for t2. And after t2 is done, you have traveled 5335 m in 90 seconds. So t1 + t2 = 90.
omg, eunice.
ReplyDeletei'm pretty sure i said the same thing as john did. and i was just saying, you only do that when you find areas. not slopes.
okay, so... i have concluded that "taek" is david. so david, thanks for #16. i get it now.
on the free response questions, for #2 part d, why is the distance 30m when you set of the equation?
ReplyDeletecan anyone teach me how to solve for two variables on TI-84 ? like, 3a. 1.7x - 2.5y = 21 ?
ReplyDeleteand...............
ReplyDeletecan someone do 1c on the pop quiz ?
and for 1b, the answer is just 5s - 6s. right ?
okay thanks Arjun I get it now. But instead of using 2 equations I just plugged in t1=90-t2 so everywhere that t1 was supposed to be I put in 90-t2 =]]
ReplyDeleteFor #3 on the review worksheet, doesn't the question mean that the dog is 21m behind and waits for 6 seconds? so that the distance between the man and the dog is 31.2m, isn't it? since 21m+(1.7m/s x 6s)=31.2m
ReplyDeleteno, the distance between the man and the dog is 21 meters no matter what. The distance the dog has to travel to a certain point, as compared with the man, is longer.
ReplyDeleteso therefore the equations would be the distance of the dog-21=distance of the man
----------------===============
ReplyDeletetaehwan, so let's say the single dotted line is 21 m and the double dotted line represents the distance the man covers. if you think about it, the entire distance (distance of the dog) minus the distance of the man is going to equal 21m.
31.2m is the distance between the man and the dog after 6 seconds, but i don't think that will help you solve the time the man gets bitten by the dog.
can someone answer my questions :(
ReplyDeleteI agree with taewhan on this one. Because it says the distance between te man and the dog is 21 meters. Then he waits 6 seconds. The man didn't stop runnig when the dog was waiting. So shouldt you have to take into account the distance he ran while the dog wa waitig. This is a poorly worded question.
ReplyDeleteAnnie, for part 1b on the pop quiz, it is t= 1-1.6ish seconds, and 5-6 seconds.
ReplyDeleteAs for part c, you just find the area above/below the lines and add/subtract accordingly to find the total displacement and where it will end up at t=7 seconds. Sorry I'm too lazy to do it all out here haha
so if I am right, then the equation would be
ReplyDelete31.2m +1.7m/s(t)= 2.5m/s(t)
and the answer would be 39s ...
thanks, tim.
ReplyDeleteso, i added all the numbers i got for area and total displacement came out to be 0.375m. is that right ? :D
OK Maggie. So whenever you have a problem which involves something being thrown up and it coming back down and you need to find the total time, you need use the displacement of the object not for d. Though it traveled up 40 and down 10, it ended up only 30 meters away from the starting point. That's where the 30 came in. The book explains it too.
ReplyDeleteYOU GUYS.
ReplyDelete#2c makes no sense at allllllllll.
"what gravity slows it down by it speed it up on the way down"
it's like, epic grammar failure.
jk, i got it.
ReplyDelete