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 some of the answer keys.
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.Good Luck!
So for number 4 on last years test, I understand the first two parts, but when I try and draw the v vs t graph, it looks nothing like the answer key. Any suggestions?
for part 3c on last year's test i know that it gives you the acceleration but what time would you use to plug into the equation in order to find the final position?
Bonjoon- The speed on the way down is 14 m/s because that is the speed you found for when the ball passes the father on the way up. Gravity slows something down (the ball going up) the same amount it speeds something up (the ball going down).
Jen- I used the eqn V^2=Vo^2 + 2ad. That way you don't have to worry about the time. You would use 0 for V because the sled is stopping, and the velocity you found in part two for Vo.
For the horse and the stuntwoman one you first find out how long it takes the woman to fall 3m then you use that to find out how far the horse runs in that time.
Okay. So I have no idea how to do number 7 on the multiple choice one. Can anyone get me started? I also don't understand how you do problems like 8,9, and 14 on the multiple choice. Is there a general rule to use for those?
Jake, the acceleration is always -9.8m/s^2. The only case where it's 0 is if the ball were at terminal velocity. there is never a case where you are throwing/dropping something that velocity and acceleration are 0.
7 is just like the Stan/Kathy problem on the other worksheet. First, make an auxiliary equation for the time, like t2+6s=t1. When the second car overtakes the first car, the distance is equal, so you plug in the numbers for the d=Vot + 1/2at^2 equation for each car, and set them equal to each other.
car1(Vot+1/2at^2)=car2(vot+1/2at^2
In order to solve for t, you plug in your auxiliary equation for t1, and then you can put it all into solver because t2 is the only variable.
1) Set up an auxiliary equation for the time; t1=t2+6 (supposing that t1 is for the first car and t2 is for the second car) 2) Set up the equation to solve for t2. I used d=Vot+.5at2, and since the point where car 2 overtakes car 1 means that they traveled the same distance, the equation would be set up as .5a1t1^2=.5a2t2^2. 3) Plug in t1=t2+6 and solve.
Kate, for #7 you set the first and second car equal to each other using the V=Vot +1/2at^2 equation. (It should look like 1/2at^2=1/2at^2) Make sure the first car is t and the second car is t-6.
For problems like 8,9 and 14, you take an equation and then see how the variable they're asking for changes when another variable doubles. For example, in 8, using the formula V=Vo=at, see what would happen to V if t is doubled. (It should double as well.)
keep in mind in saying that though nikhi if something, say a car, is moving horizontally (on the ground) or under its own power it is not necessarily affected by gravity and therefore may have an acceleration of zero
For 4c on last year's test, I was just wondering, that isn't possible in real life right? Because that would mean that the velocity somehow magically jumps to like 20m/s without accelerating at all?
Kate, to add on to what nikhi said I would suggest using some arbitrary values in the equations to prove that the conclusion you eventually come to is correct
Reema, All the term non-zero acceleration means is that it is actually accelerating, which would make the object an accelerator and after you understand that all the question is essentially asking is to describe the displacement over time graph of an accelerator which is a curved line
For #10 it helps to do a little doodley graph and work backwards - if the acceleration is constant and nonzero, then the velocity graph will be a straight diagonal line. And if the velocity v time graph is a straight diagonal line, then the position v. time graph has to be a curved line.
for 11 the velocity of the ball is zero at the highest point. therefore on the way down it must be negative velocity, pointing downward. And then gravity is always pulling downward.
For 2 on the mc, you have to first figure out the velocities for both runners by dividing 100 by their stated times. Then you multiply runner b's velocity times the time for runner a. You do this because in order to find out the distance runner b can run in runner a's time. Subtracting this number from 100 gives you how much of a lead runner b needs to tie with runner a. For 10, the displacement vz. time graph would have to be a curve since it's an accelerator. For 11, the downward means downward on the graph. Acceleration is always downward since gravity is a negative value and velocity is negative if the ball is traveling downward. For 16, at any given point, due to gravity, the speech is the same going up or down. (same question for 2b on the practice free response section)
The ball rises 40 feet and then comes down 10 feet where Joe Ro. catches the ball, meaning that he would have to be sitting 30 feet away from home plate.
oooooh that makes a lot more sense. thanks lily, I used d=VoT+.5at^2 to find the time it took for the big ball to drop 160m, and then because Tsmall=Tbig-4s, I found out the time it would take for the little ball to drop. then i used that in the d=VoT+.5at^2 equation to find out the initial velocity of the little ball.
In #5 on the MC, why can an object have increasing speed while the acceleration is decreasing? I'm having a hard time thinking of an example of that situation that I can understand
Alright Asha. Here's an example for you. I hope it makes sense :D #5. Let's say you are holding a block in your hands. Air friction isn't affecting the block, but gravity is pushing it down. Now when you drop the block, the first thing that affects it is gravity. Then air friction comes into play and slows the acceleration down. Gravity is still pushing and it hits the floor. In this case the acceleration is decreasing but the object is still increasing in speed.
If that doesn't make sense, try to graph it out. Even as the acceleration slows down, velocity is still going up, just at a slower rate.
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!
So for number 4 on last years test, I understand the first two parts, but when I try and draw the v vs t graph, it looks nothing like the answer key. Any suggestions?
ReplyDeleteAsha- Did your slopes match up with the ones on the answer key? (x,y) coordinates for v vs t graph are (time, d vs t slope)
ReplyDeleteOk, that makes sense. I think I was getting the types of graphs confused lol. Thanks Jackie!
ReplyDeletefor #2 on last year's test, I don't get part c. how can the speed on the way down be 14m/s ?
ReplyDeleteHow did Mr. Rodino solve #15 on the multiple choice worksheet from last night
ReplyDeleteBrian- he used V=Vo+at but V= -Vo because it comes down at the same speed it was shot up at.. So the eqn becomes -Vo = Vo+-9.8(10s)
ReplyDeletefor part 3c on last year's test i know that it gives you the acceleration but what time would you use to plug into the equation in order to find the final position?
ReplyDeleteBonjoon-
ReplyDeleteThe speed on the way down is 14 m/s because that is the speed you found for when the ball passes the father on the way up. Gravity slows something down (the ball going up) the same amount it speeds something up (the ball going down).
Jen-
I used the eqn V^2=Vo^2 + 2ad. That way you don't have to worry about the time. You would use 0 for V because the sled is stopping, and the velocity you found in part two for Vo.
can anyone tell me what needs to be calculated for the hors in the question with the horse and stuntwoman?
ReplyDeleteKatsuya -
ReplyDeleteFor the horse and the stuntwoman one you first find out how long it takes the woman to fall 3m then you use that to find out how far the horse runs in that time.
Okay. So I have no idea how to do number 7 on the multiple choice one. Can anyone get me started? I also don't understand how you do problems like 8,9, and 14 on the multiple choice. Is there a general rule to use for those?
ReplyDeleteFor number 12 on the MC, why is the answer b. If at the highest point the ball is not moving, shouldn't the velocity and acceraltion be zero?
ReplyDeleteJake,
ReplyDeletethe acceleration is always -9.8m/s^2. The only case where it's 0 is if the ball were at terminal velocity. there is never a case where you are throwing/dropping something that velocity and acceleration are 0.
Reynolds-
ReplyDelete7 is just like the Stan/Kathy problem on the other worksheet.
First, make an auxiliary equation for the time, like t2+6s=t1. When the second car overtakes the first car, the distance is equal, so you plug in the numbers for the d=Vot + 1/2at^2 equation for each car, and set them equal to each other.
car1(Vot+1/2at^2)=car2(vot+1/2at^2
In order to solve for t, you plug in your auxiliary equation for t1, and then you can put it all into solver because t2 is the only variable.
Re: Kate R's question (#7 on MC)
ReplyDelete1) Set up an auxiliary equation for the time; t1=t2+6 (supposing that t1 is for the first car and t2 is for the second car)
2) Set up the equation to solve for t2. I used d=Vot+.5at2, and since the point where car 2 overtakes car 1 means that they traveled the same distance, the equation would be set up as .5a1t1^2=.5a2t2^2.
3) Plug in t1=t2+6 and solve.
-Christine Q
Kate, for #7 you set the first and second car equal to each other using the V=Vot +1/2at^2 equation. (It should look like 1/2at^2=1/2at^2) Make sure the first car is t and the second car is t-6.
ReplyDeleteFor problems like 8,9 and 14, you take an equation and then see how the variable they're asking for changes when another variable doubles. For example, in 8, using the formula V=Vo=at, see what would happen to V if t is doubled. (It should double as well.)
Jake, gravity is always present so therefore, acceleration should not be zero.
ReplyDeletetoo much work without a calc
ReplyDeletekeep in mind in saying that though nikhi if something, say a car, is moving horizontally (on the ground) or under its own power it is not necessarily affected by gravity and therefore may have an acceleration of zero
ReplyDeleteyea i think a speeder has acc. of 0
ReplyDeleteFor 4c on last year's test, I was just wondering, that isn't possible in real life right? Because that would mean that the velocity somehow magically jumps to like 20m/s without accelerating at all?
ReplyDeleteCan someone please explain # 10 in the multiple choice to me?
ReplyDeleteKate,
ReplyDeleteto add on to what nikhi said I would suggest using some arbitrary values in the equations to prove that the conclusion you eventually come to is correct
for number 11 on the mc what does it mean by it points upward or downward
ReplyDeleteReema,
ReplyDeleteAll the term non-zero acceleration means is that it is actually accelerating, which would make the object an accelerator and after you understand that all the question is essentially asking is to describe the displacement over time graph of an accelerator which is a curved line
for #15 on the mc i got 98m/s but the answer is 49m/s why do you cut it in half?
ReplyDeleteOhh okayy. thankss.
ReplyDeletenvm i got it
ReplyDeleteanyone feel like explaining numbers 2,10,11,16 on the m/c
ReplyDeleteFor #10 it helps to do a little doodley graph and work backwards - if the acceleration is constant and nonzero, then the velocity graph will be a straight diagonal line. And if the velocity v time graph is a straight diagonal line, then the position v. time graph has to be a curved line.
ReplyDeletefor 11 the velocity of the ball is zero at the highest point. therefore on the way down it must be negative velocity, pointing downward. And then gravity is always pulling downward.
ok thanks for the help i guess i cant read
ReplyDeleteFor 2 on the mc, you have to first figure out the velocities for both runners by dividing 100 by their stated times. Then you multiply runner b's velocity times the time for runner a. You do this because in order to find out the distance runner b can run in runner a's time. Subtracting this number from 100 gives you how much of a lead runner b needs to tie with runner a.
ReplyDeleteFor 10, the displacement vz. time graph would have to be a curve since it's an accelerator.
For 11, the downward means downward on the graph. Acceleration is always downward since gravity is a negative value and velocity is negative if the ball is traveling downward.
For 16, at any given point, due to gravity, the speech is the same going up or down. (same question for 2b on the practice free response section)
For 2b in last years test review, i dont understand why we input d as 30m in the equation. can someone exaplin that to me?
ReplyDeleteAsian Dan-
ReplyDeleteThe ball rises 40 feet and then comes down 10 feet where Joe Ro. catches the ball, meaning that he would have to be sitting 30 feet away from home plate.
For the ratio problems, how do you decide which equation to use? can you just use any that have the givens as variables?
ReplyDeletecan anyone help me with 8 on the multiple choice? which equation do you use?
ReplyDeleteellie,
ReplyDeleteI used v = vo + at. You don't have to rearrange it or anything because the problem's asking for the final v.
for 3a on the review problems sheet with the one about the attack dog, the dog is an accelerator, right?
ReplyDeleteMegan,
ReplyDeleteThe dog is not an accelerator. His average velocity is high enough to bite the dude without accelerating.
-Christine
Hey guys, for 1c) on the 1D Motion 2008 review problems page, what did you guys get for the velocity? and how did you approach the problem?
ReplyDeleteoooooh that makes a lot more sense. thanks
ReplyDeletelily,
I used d=VoT+.5at^2 to find the time it took for the big ball to drop 160m, and then because Tsmall=Tbig-4s, I found out the time it would take for the little ball to drop. then i used that in the d=VoT+.5at^2 equation to find out the initial velocity of the little ball.
In #5 on the MC, why can an object have increasing speed while the acceleration is decreasing? I'm having a hard time thinking of an example of that situation that I can understand
ReplyDeleteAlright Asha.
ReplyDeleteHere's an example for you. I hope it makes sense :D
#5. Let's say you are holding a block in your hands. Air friction isn't affecting the block, but gravity is pushing it down.
Now when you drop the block, the first thing that affects it is gravity. Then air friction comes into play and slows the acceleration down. Gravity is still pushing and it hits the floor. In this case the acceleration is decreasing but the object is still increasing in speed.
If that doesn't make sense, try to graph it out. Even as the acceleration slows down, velocity is still going up, just at a slower rate.
Does that make sense?
yea, it does. thanks!
ReplyDeletehey everyone, good luck on the test tomorrow!!
ReplyDeleteI am not quite understanding the explanation for 2c. Can anyone elaborate upon what already is on the answer key?
ReplyDeletei think it means that gravity acts on the ball on the way up and when it is coming down
ReplyDeletei also tried calc v for up and down and they both came out the same, 14m/s
ABANDONED!!! poor blog... *hugs*
ReplyDelete