Same assignment...post a question or answer one for your points. Here are the keys in case you could access them. Good luck sudying!! Look at 1st period for the answer keys in case you cannot find them.
can someone help me for number 3 in the kinematics and graphing packet?? the equation that i used was d = Vot + 1/2at^2. i know that you have to at the 2 differnt times together and equal it to 5334. idk what to do after.
LAUREN: 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?
Arjun Arya said...
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.
If anyone is having trouble forming the INITIAL equation, you use d1+d2= 5334, and now use the FIRST equation to get the initial equation. Remember, the initial velocity for D1 is ZERO, and the ACCELERATION for D2 is ZERO.
can someone explain to me how you do #3 on 1d motion practice problems 2? and for 4b on the same worksheet i got the right answer, but.. how can the velocity be negative? because if the keys were thrown UP then shouldn't the velocity be positive?
-on D v. T graphs slope equals average velocity. -on V v. T graphs slope = average acceleration -on A v. T graphs the lines are always horizontal. -Find area on V v. T graphs in order to graph displacement vs. Time -
#3 on the review problems i did: dog's initial velocity*dog's time-man's initial velocity*man's time=21m and when you plug in the numbers you get: 2.5m/s*t-1.7m/s*t=21m and when i solved for t i got the man's time to be 26.25s & and dog's time to be 20.25s but mr. rodino's answer key says it's 45s & 39s and we did the exact same steps.. plugged in the exact same numbers... where did i mess up???!
CJ - i think you might be reffering to the d=vt? you use this equation to sub in vt when you do not have d and are looking for another variable in the eqn.
this also works to replace v and t in similar situations if you solve the equation for whatever variable you'd like to sub the other two variables in for.
Ok. For the graphs that we did early in the year, all we need to use are tangent lines to check if our displacement, velocity, and acceleration graphs connect, right? If I'm wrong about anything please correct me guys.
About terminal velocity, i think we only need to know that it means zero acceleration. Mr. Rodino said in class today that we would not have to do any calculations with air friction. we only need to know the concept, which is that air friction makes it possible for speed to increase while acceleration decreases
THIS IS JAMIE NOT JAMES hahahaha dude for a sec i thought mike h was mike hsu!!
by the way, why are u posting in this arjun? just wondering...
for 3.a on last year's test i got the same answer as arjun, t1 = 5 sec and t2 = 85s, but if u add it together the total distance is only like 1000m or so not 5334m... what am i doing wrong?
CJ, i think you're talking about the d=vt equation. someone brought it up but just to add on, you only use this equation when doing interval problems. example problem is #2 on worksheet 1D motion practice problems 1. just a side note, when velocity is constant it's average velocity is the same
How again do you do that problem on the mc with the car going at a velocity of 30 m/s then stopping at distance d then how long would it take to stop if it were going two times faster... i remember something about some funky symbol, but other than that i forgot...
SOMEONE. ANYONE. PLEASE. explain to me how you do #3 on 1d motion practice problems 2? and for 4b on the same worksheet i got the right answer, but.. how can the velocity be negative? because if the keys were thrown UP then shouldn't the velocity be positive?
Ok, really simple question, ANSWER ME!!! Do you need to have the Ti 89's solver to do #3 on last year's test?? I dony see how to plug it into the ti 84 and my equations are right.
to double check it: for the first part you use d=vot+(1/2)(a)(t1) squared initial velocity=0 so... d= (1/2)(12.2)(25)=152.5m and then you find out the velocity when the engine turns off using v^2=vo^2 + 2ad and you get v^2=2(12.2)(152.5)=61m/s
and then use d=v(t2) to find out the second distance... d=(61)(85)=5185m
5185+152.5= 5337.5 pretty close to 5334m so i think 5s and 85s is the right answer.
no they are all optional i suggest doing them though, they really helped me a lot. thanks rosako anyways.
ok andy u prolly cant do it on ur weak little machine u probably need the ti 89, but what u can do is set it to 0, and combine like terms, then u can use POLYSMLT
Alright first comment!
ReplyDeleteGood luck on the test tomorrow!
and the acceleration of gravity is -9.8m/s^2
How do solve for t1 and t2 in problem 3 of last years test?
ReplyDeleteHey moez thanks for the help on the problems earlier today. They were difficult because I was doing the wrong ones
ReplyDeletecan someone help me for number 3 in the kinematics and graphing packet?? the equation that i used was d = Vot + 1/2at^2.
ReplyDeletei know that you have to at the 2 differnt times together and equal it to 5334. idk what to do after.
FOR EVERYONE HAVING TROUBLE ON THREE:
ReplyDeleteLAUREN: 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?
Arjun Arya said...
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.
If anyone is having trouble forming the INITIAL equation, you use d1+d2= 5334, and now use the FIRST equation to get the initial equation. Remember, the initial velocity for D1 is ZERO, and the ACCELERATION for D2 is ZERO.
Good Luck!
can someone explain to me how you do #3 on 1d motion practice problems 2? and for 4b on the same worksheet i got the right answer, but.. how can the velocity be negative? because if the keys were thrown UP then shouldn't the velocity be positive?
ReplyDeleteGood luck on the test guys.
ReplyDeleteSome quick things to remember:
-on D v. T graphs slope equals average velocity.
-on V v. T graphs slope = average acceleration
-on A v. T graphs the lines are always horizontal.
-Find area on V v. T graphs in order to graph displacement vs. Time
-
How would set up problem 15 in the multiple choice packet>
ReplyDelete#3 on the review problems
ReplyDeletei did:
dog's initial velocity*dog's time-man's initial velocity*man's time=21m
and when you plug in the numbers you get:
2.5m/s*t-1.7m/s*t=21m
and when i solved for t i got the man's time to be 26.25s & and dog's time to be 20.25s
but mr. rodino's answer key says it's 45s & 39s
and we did the exact same steps.. plugged in the exact same numbers...
where did i mess up???!
For your second t it should be t + 6
ReplyDeleteMr. Rodino's paper says
(2.5m/s)TD - (1.7)TM = 21
TM and TD are not the same TM = TD + 6 seconds
if you plug that in it will work out correctly
How do i put two equations into my calculator and make it look for one answer? i have the 89
ReplyDeleteI remember one of the equations we learned had to be used in only a certain scenario. Does anyone know which I am referring to and when to use it?
ReplyDeletehow do you deal with air friction in a word probelm?
ReplyDeleteNo idea. what were we talking about?
ReplyDeletekelsey i believe you use gravity till it hits terminal velocity
ReplyDeleteCJ - i think you might be reffering to the d=vt?
ReplyDeleteyou use this equation to sub in vt when you do not have d and are looking for another variable in the eqn.
this also works to replace v and t in similar situations if you solve the equation for whatever variable you'd like to sub the other two variables in for.
hope this is the one you were talking about
mike h - when you say "terminal velocity", is that value given to us in the word problem? or do we have to find it on our own?
ReplyDeletedavid, i did that but do you put it into solver?? or not??
ReplyDeleteoh. never mind, i got it.
ReplyDeletewait how do you find the terminal velocity???
ReplyDeleteyeah that's what i'm asking too james
ReplyDeletelol
btw your last name is my middle name
to answer kmukhi_93's question for number 15 on MC, you use the equation
ReplyDelete(0m/s)=Vnot + (-9.8m/s^2)(5s)
Hey guys good luck tomorrow.
ReplyDeleteOk. For the graphs that we did early in the year, all we need to use are tangent lines to check if our displacement, velocity, and acceleration graphs connect, right? If I'm wrong about anything please correct me guys.
Thanks!!!!!!
About terminal velocity, i think we only need to know that it means zero acceleration.
ReplyDeleteMr. Rodino said in class today that we would not have to do any calculations with air friction.
we only need to know the concept, which is that air friction makes it possible for speed to increase while acceleration decreases
Thanks Arjun Arya!
ReplyDeleteI finally could solve the question #3 :):):)
Thank god Elif...
ReplyDeleteJames which equation is that?? Why is the -9.8 squared? Im lost..
hahahhahh. i love everyone's comments after someone answered a question.
ReplyDeleteahahhahah and thanks elif for the explaination.
LOLOL
THIS IS JAMIE NOT JAMES
ReplyDeletehahahaha dude for a sec i thought mike h was mike hsu!!
by the way, why are u posting in this arjun? just wondering...
for 3.a on last year's test i got the same answer as arjun, t1 = 5 sec and t2 = 85s, but if u add it together the total distance is only like 1000m or so not 5334m... what am i doing wrong?
oh nevermind i got it
ReplyDeleteCJ, i think you're talking about the d=vt equation.
ReplyDeletesomeone brought it up but just to add on, you only use this equation when doing interval problems. example problem is #2 on worksheet 1D motion practice problems 1.
just a side note, when velocity is constant it's average velocity is the same
How again do you do that problem on the mc with the car going at a velocity of 30 m/s then stopping at distance d then how long would it take to stop if it were going two times faster... i remember something about some funky symbol, but other than that i forgot...
ReplyDeletehow do you use solver on ti 84?
ReplyDeleteSOMEONE. ANYONE. PLEASE. explain to me how you do #3 on 1d motion practice problems 2? and for 4b on the same worksheet i got the right answer, but.. how can the velocity be negative? because if the keys were thrown UP then shouldn't the velocity be positive?
ReplyDeleteDoes it help to have a ti 89 for #3 on last years test?
ReplyDeleteno, you don't need an 89 if u have polysmlt on ur regular calculator. if u dont, solver pretty much helps on every problem
ReplyDeletejohn,
ReplyDeletemath --> solver --> enter --> plug in equation (make sure it is equal to 0) --> hit enter --> punch in the number 10 --> alpha --> solve
what concepts do we review for this test?
ReplyDeleted v. t graphs
v. v. t graphs
a v. t graphs
formulas
what else did we learn?
Ok, really simple question, ANSWER ME!!! Do you need to have the Ti 89's solver to do #3 on last year's test?? I dony see how to plug it into the ti 84 and my equations are right.
ReplyDeletedony= dont
ReplyDeleteto double check it:
ReplyDeletefor the first part you use d=vot+(1/2)(a)(t1) squared
initial velocity=0
so... d= (1/2)(12.2)(25)=152.5m
and then you find out the velocity when the engine turns off using v^2=vo^2 + 2ad and you get
v^2=2(12.2)(152.5)=61m/s
and then use d=v(t2) to find out the second distance... d=(61)(85)=5185m
5185+152.5= 5337.5 pretty close to 5334m so i think 5s and 85s is the right answer.
oh jk i was answering jamies question i didnt see that he said he got it...
ReplyDeletedoes anyone know if we have any ws that are actually DUE tomorrow? or are they all just optional?
ReplyDeleteno they are all optional i suggest doing them though, they really helped me a lot. thanks rosako anyways.
ReplyDeleteok andy u prolly cant do it on ur weak little machine u probably need the ti 89, but what u can do is set it to 0, and combine like terms, then u can use POLYSMLT
freaking out rt now :(
ReplyDeletedo we need to know that fourth equation for the test? i know we need those main three, and avg. velocity, but is that it?
ReplyDeletei think so, though the d=vt might be helpful as well
ReplyDeleteThe best study tip, I can think of is as long as you know all the equations, and substitute often, you should be golden.
ReplyDelete