Remember that you have to make one post either a question or an answer to a questions. Please keep comments physics related...I will be reading these. Fell free to answer mulitple times.
Hey Kyle! For the third part (whoa, 9:05 AM?) you had to use the V^2= Vo^2 + 2ad since you had the inital velocity, the final velocity (0 because when your at the top, there's a moment of no movement) and then since after the acceleration of the engines stopped, gravity would take over! So all you had to do was solve for D! And then you added it to the other distances you calculated. Our group got somewhere around 161.1 i think? Hope i helped.!
Why can an object have increasing speed while its acceleration is decreasing? (Practice multiple choice #5) I got the answer right but I'm not exactly sure I understand the logic behind it.
Your in a car at rest, and you SLAM THE PETAL right? Now your car is starting at its maximum acceleration, as the car continues to go faster, you'll relax your foot off the petal, and your acceleration if now decreasing. Yet, as you got faster, you let the gas petal go a little by little. So, as you got faster, your acceleration was decreasing.
Anne, if an object has a negative velocity and a negative acceleration, it's still speeding up. Or at least, if increasing speed means speeding up haha.
Tim, if an object has neg. velocity and negative accel., it is simply speeding up while traveling away (negatively) from the origin. But, that would simply mean acceleration is constant no?
If your velocity is negatively changing at (-1 m/s) then you want to DECREASE your acceleration, meaning if your negative, it would be going closer to zero, you would have to let air friction take over. As in the car example, if you are going 60 mph, and you take your foot off the petal, your acceleration is now decreasing, but the force of air is still propelling you forward at 60 mph for a few seconds before you actually start slowing down.
And Arjun, think about it like this V^2 = Vo^2 + 2ad, 30^2 = V0^2+ 2ad 2(30^2)= 2(Vo^2 + 2ad) thus resulting in 4d. I dont know if this helps but it helped me
Yes, you can use a calculator, but I advise only using it when you absolutely cannot do it in your head. Mr. Rodino said that during multiple choice part of our test we will not be able to use our calculators, so yup.
for question number 3 on the practice problems, what would be your initial equation? I think you would have to solve for the distance both of them go to be equal so you'd set the ball dropped equal to the ball thrown instead= to d, but im not sure that would get me anywhere
Oh, and Anne, another way to look at the acceleration problem:
Imagine a car that goes 0 to 60 m/s in 5 seconds. This same car has a top speed of 120 m/s.
If I slam the pedal, it's certainly going to take longer to go from 60 to 120 than it did to go from 0 to 60, and by the time I hit 120 my acceleration will be at 0. During the whole time, my acceleration was decreasing but my velocity continued to increase.
Tim and Anne, an easier example would be to imagine a rocket with an engine. It starts not moving and then blasts off at a major acceleration because of the rocket fuel and stuff. And when the rocket fuel runs out, and the rocket starts to fall, there will be a point, right as the engine runs out of fuel, that the rocket will be going faster (air friction), and the acceleration will slowly be decreasing to -9.8m/s. THEN, once the acceleration is negative, the rocket will start to fall.
And Anne for #6, the brakeing force simply means you are applying constant (negative) pressure, the number doesn't really matter. And twice as fast is yes, 60m/s.
Ok Arjun, I don't know if my first way made no sense so here is another way to explain it When it says same breaking force it means that it is the same deceleration rate as when the car was going 30 m/s. So what that means is when even though your going 60 m/s, you are still decelerating at the same rate. Normally, If you double speed, everything (including acceleration and distance) will double. However, you are not doubling acceleration. to make up for that, you quadruple the distance thus making it 4d. Again I suck at explaining so if this makes no sense sorry.
Sid, i still don't get it, lol. There must be some use of some eqn, the eqn's you used above can't exactly be used because we don't know initial velocity.?
oh wait, lol i Meant to say Anne,or who ever asked about number 6. Ya the with the equations Im just trying to say that velocity is doubled and distance is too but acceleration remains the same so you double distance again. I don't know if that makes sense either but it does to me.
this has nothing to do with the work sheet, but can someone expain to me how area and slope and positives and negatives work into converting a v vs. t graph to both an a vs. t graph and d vs. t graph?
Emily, for number 3, since i haven't done it, i do not know the exact equation. BUT, think of it this way, then the two balls meet, their TIME will be the same. Use that information to come up with 2 equations since you know initial velocity and stuff and set those equals and yeah.
First, you want to set the distances = to each other (The balls will not have traveled the same distance, but they will be the same distance from the ground. So we will write the equation using the ground as our x-axis).
The Left side is the ball that is thrown upward (hence the 25*t for initial velocity), the right side is the dropped ball (you add 15m because it starts that far off the ground).
Maggie, from a v vs. t graph to an a vs. t, you use the slopes to determine whether the lines go above, below, or on the x-axis. From a v vs. t to a d vs. t, that is when you use the areas above and below the x axis to create the points on the d vs. t graph. Also, wherever the line is horizontal on the v vs. t, you draw a straight line in between the 2 points on the d vs. t. yea idk if that answered you're question at all but whatever haha
On Multiple choice question #6 with the car traveling 30 m/s and stopping in distance d. With the car doubling the speed it stops in distance 4d. I got the right answer because i've seen these questions before, but i still have no idea how that's correct. Can someone please explain that to me.
I'm with John. For Number 6 MC, i got it right because i know that as velocity increases, you have to square the distance for that and stuff thanks to Drivers Ed. (Woohoo?) But, i don't get it, would someone use equations or something to solve it? Thanks.
Yeah sid tried to explain to me #6 earlier (like what he said above) but that went over my head because if you use the equation : {v = vot + 2ad} then wouldnt the stopping time for #1 be 2d with that logic?
Okay, I tried to look for David's entry because lalbert said his explanation helped, but I don't see any David's here... so I'm pretty frustrated right now. :(
Uhm, but Taek, the acceleration in your equation would be -9.8m/s^2. right ? Not m/s ?
For #4b, I got a negative velocity. Is that right ? If it is, I don't understand why because I thought negative velocity meant going towards where you started.
Ok let me try again with equations V^2= Vo + 2ad V^2= 0, Vo=30 m/s 0=30 m/s+ 2ad -30 m/s = 2ad 2(-30 m/s)= 2(2ad) BUT REMEMBER, A IS NOT BEING MULTIPLIED BY 2 IN THIS SCENARIO, IT REMAINS THE SAME. so the 2 only applies to the d so therefore -60 m/s = a times 4d. Ok this is this closest i can get to explaining it with equations. Yes I know this sucks lol.
Sid, I don't know if this changes anything in what you did but it should be V not V^2 When i tried solved it, i used the equations V= Vo + 2ad1 and V= 2Vo + 2ad2 I set V= 0, Vo=30 m/s, and A= to -9.8 m/s^2 Since V=0 in both i set the equations equal to each other and got Vo+2ad1=2Vo+2ad2 I the closest i could get was d1+d2=Vo/(19.6m/s^2) I wasnt sure where to go from here
Sid and justin. I dont know it this could help at all but i think what were doing is similar to questions 8 and 9. See, the object would be going twice as fast but if its constant it would be 4 times as far. I think that may help all the confusion and writing down equations and stuff.
Nevermind what I said, I was looking at a different problem. For number 6, sid, since the equation for velocity with distance is V^2=Vo^2+2ad, in both instances the 2a cancels out since the braking speed is consistent. It's 4 times as large because when you square a value that is twice as large, it becomes four times bigger, (ie 4^2=16 and 8^2=64). So that is why the answer is 4d
i used the first equation, too. for 4a, that is. i didn't get an old answer for 4a, but for 4b, i used the third equation (which is the only equation you can use out of the three), i got a negative velocity which doesn't make sense.
annie i just looked @ the problem and i got a negative and positive answer, same number different signs. easily its the positive answer. idk what you did wrong. i used equation #2 which worked better for me since you know everything in that eqn except the velocity.
if you did #5b, tell me if i did it right. i set up an equation for the stuntwoman by using V=Vo + at. then, i set up an equation for the horse by using d = Vot + 1/2at^2. then i made them equal to each other and found t.
hmm for #5b its fairly simple. i did d=vot + 1/2at^2 because she falls there is no inital velocity so it ends up turning into d = 1/2 at^2 you know every variable on that chart except time so its pretty easy, no need for 2 equations. also dont forget she is falling DOWN!
I didn't quite understand what #3 was asking. Who's supposed to win ? - - because it says 'largest lead the team B runnner can have...' and 'in order that team A runner not lose the race'
How do you do the 3rd part to the rocket group problem and what formula what you use for that part?
ReplyDeleteHey Kyle! For the third part (whoa, 9:05 AM?) you had to use the V^2= Vo^2 + 2ad since you had the inital velocity, the final velocity (0 because when your at the top, there's a moment of no movement) and then since after the acceleration of the engines stopped, gravity would take over! So all you had to do was solve for D! And then you added it to the other distances you calculated. Our group got somewhere around 161.1 i think? Hope i helped.!
ReplyDeleteWhy can an object have increasing speed while its acceleration is decreasing? (Practice multiple choice #5) I got the answer right but I'm not exactly sure I understand the logic behind it.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThe Answer is Yes BECAUSE, think of it this way.
ReplyDeleteYour in a car at rest, and you SLAM THE PETAL right? Now your car is starting at its maximum acceleration, as the car continues to go faster, you'll relax your foot off the petal, and your acceleration if now decreasing. Yet, as you got faster, you let the gas petal go a little by little. So, as you got faster, your acceleration was decreasing.
It made sense to me :P
Anne, if an object has a negative velocity and a negative acceleration, it's still speeding up. Or at least, if increasing speed means speeding up haha.
ReplyDeleteTim, if an object has neg. velocity and negative accel., it is simply speeding up while traveling away (negatively) from the origin. But, that would simply mean acceleration is constant no?
ReplyDeleteIf your velocity is negatively changing at (-1 m/s) then you want to DECREASE your acceleration, meaning if your negative, it would be going closer to zero, you would have to let air friction take over. As in the car example, if you are going 60 mph, and you take your foot off the petal, your acceleration is now decreasing, but the force of air is still propelling you forward at 60 mph for a few seconds before you actually start slowing down.
Thanks Tim! Arjun?? lol
ReplyDeleteI have another question: (Practice Multiple Choice #6)"Assuming the same braking force" does this mean acceleration = -30m/s^2.
Also, "traveling twice as fast" does this mean that velocity = 60m/s?
OK this isn't really a question about a specific problem but are we aloud to use calculator on this multiple choice?
ReplyDeleteAnd Arjun, think about it like this
ReplyDeleteV^2 = Vo^2 + 2ad,
30^2 = V0^2+ 2ad
2(30^2)= 2(Vo^2 + 2ad) thus resulting in 4d. I dont know if this helps but it helped me
This multiple choice is from the book, so yes.
ReplyDeleteThe test multiple choice, no.
(umm, this is David Vortex. My google account just happens to be my online alias)
Read the directions =]
ReplyDeleteYes, you can use a calculator, but I advise only using it when you absolutely cannot do it in your head. Mr. Rodino said that during multiple choice part of our test we will not be able to use our calculators, so yup.
for question number 3 on the practice problems, what would be your initial equation? I think you would have to solve for the distance both of them go to be equal so you'd set the ball dropped equal to the ball thrown instead= to d, but im not sure that would get me anywhere
ReplyDeleteOh, and Anne, another way to look at the acceleration problem:
ReplyDeleteImagine a car that goes 0 to 60 m/s in 5 seconds.
This same car has a top speed of 120 m/s.
If I slam the pedal, it's certainly going to take longer to go from 60 to 120 than it did to go from 0 to 60, and by the time I hit 120 my acceleration will be at 0. During the whole time, my acceleration was decreasing but my velocity continued to increase.
Tim and Anne, an easier example would be to imagine a rocket with an engine. It starts not moving and then blasts off at a major acceleration because of the rocket fuel and stuff. And when the rocket fuel runs out, and the rocket starts to fall, there will be a point, right as the engine runs out of fuel, that the rocket will be going faster (air friction), and the acceleration will slowly be decreasing to -9.8m/s. THEN, once the acceleration is negative, the rocket will start to fall.
ReplyDeleteAnd Anne for #6, the brakeing force simply means you are applying constant (negative) pressure, the number doesn't really matter. And twice as fast is yes, 60m/s.
David, that helped a lot. Thankss!
ReplyDelete(Multiple Choice #6): scratch what i said earlier. how come the answer is 4d and not 2d?
Ok Arjun, I don't know if my first way made no sense so here is another way to explain it
ReplyDeleteWhen it says same breaking force it means that it is the same deceleration rate as when the car was going 30 m/s. So what that means is when even though your going 60 m/s, you are still decelerating at the same rate. Normally, If you double speed, everything (including acceleration and distance) will double. However, you are not doubling acceleration. to make up for that, you quadruple the distance thus making it 4d. Again I suck at explaining so if this makes no sense sorry.
Sid, i still don't get it, lol. There must be some use of some eqn, the eqn's you used above can't exactly be used because we don't know initial velocity.?
ReplyDeleteoh wait, lol i Meant to say Anne,or who ever asked about number 6. Ya the with the equations Im just trying to say that velocity is doubled and distance is too but acceleration remains the same so you double distance again. I don't know if that makes sense either but it does to me.
ReplyDeletethis has nothing to do with the work sheet, but can someone expain to me how area and slope and positives and negatives work into converting a v vs. t graph to both an a vs. t graph and d vs. t graph?
ReplyDeleteArjun, I'll show you what I mean tomorrow in class. Hard to explain it since it's a diagram from my calc notebook haha
ReplyDeleteCan someone help me with my question.... Thanks :)
ReplyDeleteEmily, for number 3, since i haven't done it, i do not know the exact equation. BUT, think of it this way, then the two balls meet, their TIME will be the same. Use that information to come up with 2 equations since you know initial velocity and stuff and set those equals and yeah.
ReplyDeleteGood luck ;)
Emily, I have the Q in front of me.
ReplyDeleteFirst, you want to set the distances = to each other (The balls will not have traveled the same distance, but they will be the same distance from the ground. So we will write the equation using the ground as our x-axis).
25m/s(t) + 1/2(-9.8m/s)(t)^2 = 15m + 1/2(-9.8m/s)t^2)
The Left side is the ball that is thrown upward (hence the 25*t for initial velocity), the right side is the dropped ball (you add 15m because it starts that far off the ground).
Maggie, from a v vs. t graph to an a vs. t, you use the slopes to determine whether the lines go above, below, or on the x-axis. From a v vs. t to a d vs. t, that is when you use the areas above and below the x axis to create the points on the d vs. t graph. Also, wherever the line is horizontal on the v vs. t, you draw a straight line in between the 2 points on the d vs. t. yea idk if that answered you're question at all but whatever haha
ReplyDeleteDavid your explaination for #3 really helped. Can someone tell me how to approach #5??
ReplyDeleteOn Multiple choice question #6 with the car traveling 30 m/s and stopping in distance d. With the car doubling the speed it stops in distance 4d. I got the right answer because i've seen these questions before, but i still have no idea how that's correct. Can someone please explain that to me.
ReplyDeleteActually David where did you get that 15 on the right side?
ReplyDeleteThe 15 comes from the ball starting 15m high.
ReplyDeleteFor #5, try B first and work backwards.
ReplyDeletebut isn't the equation supposed to be initial velocity times time plus 1/2 at?
ReplyDeleteFor numer three btw
ReplyDeleteI'm with John. For Number 6 MC, i got it right because i know that as velocity increases, you have to square the distance for that and stuff thanks to Drivers Ed. (Woohoo?) But, i don't get it, would someone use equations or something to solve it? Thanks.
ReplyDeleteYeah sid tried to explain to me #6 earlier (like what he said above) but that went over my head because if you use the equation : {v = vot + 2ad} then wouldnt the stopping time for #1 be 2d with that logic?
ReplyDeletesorry i meant for the first speed not #1
ReplyDeleteFor number three we are looking at the distance to get the time.
ReplyDeleteD = Vo (which is 0) + .5(a)t^2
Okay, I tried to look for David's entry because lalbert said his explanation helped, but I don't see any David's here... so I'm pretty frustrated right now. :(
ReplyDeleteUhm, but Taek, the acceleration in your equation would be -9.8m/s^2. right ? Not m/s ?
For #4b, I got a negative velocity. Is that right ? If it is, I don't understand why because I thought negative velocity meant going towards where you started.
Okay if that's the case why is the right side of the equation15 +. 1/2(-9.8)(t^2) why does 15 go there shouldn't it be 0?
ReplyDeletemaybe..............
ReplyDeletejust maybe.........
the 15 m goes on the right side because it's 15 meters high and d=vt. so... the Vot is replaced with d ?
That could make sense but I thought initial velocity was 0
ReplyDeletea general question..but is the position vs. time graph the same as distance vs. time?
ReplyDeleteHey guys and gals, how did you solve number 15 on the mc? I'm stumped
ReplyDeleteyeah, man.
ReplyDeletei think.
maybe it's displacement vs time ?
is there such thing...?
everyone left. i don't think they're coming back..
Ok let me try again with equations
ReplyDeleteV^2= Vo + 2ad
V^2= 0, Vo=30 m/s
0=30 m/s+ 2ad
-30 m/s = 2ad
2(-30 m/s)= 2(2ad)
BUT REMEMBER, A IS NOT BEING MULTIPLIED BY 2 IN THIS SCENARIO, IT REMAINS THE SAME. so the 2 only applies to the d
so therefore
-60 m/s = a times 4d. Ok this is this closest i can get to explaining it with equations. Yes I know this sucks lol.
lol, well sid i think i get it a bit better now that you did all the math. thanks a bunch man.
ReplyDeleteok, no one want to ask more questions i can answer to reassure myself that i actually know this stuff? lol.
ReplyDeleteSid, I don't know if this changes anything in what you did but it should be V not V^2
ReplyDeleteWhen i tried solved it, i used the equations
V= Vo + 2ad1 and V= 2Vo + 2ad2
I set V= 0, Vo=30 m/s, and A= to -9.8 m/s^2
Since V=0 in both i set the equations equal to each other and got Vo+2ad1=2Vo+2ad2
I the closest i could get was
d1+d2=Vo/(19.6m/s^2)
I wasnt sure where to go from here
Sid and justin. I dont know it this could help at all but i think what were doing is similar to questions 8 and 9. See, the object would be going twice as fast but if its constant it would be 4 times as far. I think that may help all the confusion and writing down equations and stuff.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteNevermind what I said, I was looking at a different problem. For number 6, sid, since the equation for velocity with distance is
ReplyDeleteV^2=Vo^2+2ad, in both instances the 2a cancels out since the braking speed is consistent. It's 4 times as large because when you square a value that is twice as large, it becomes four times bigger, (ie 4^2=16 and 8^2=64). So that is why the answer is 4d
wow, justin that made a lot of sense. thanks.
ReplyDeleteHow would you solve problem 4 on the practice problems? I used equation number one of the gravity equations, but I'm getting a really odd answer.
ReplyDeleteYOU GUYS.
ReplyDeletedid you get a negative velocity for 4b ?
i asked earlier, but no one answered.
this sucks.
eunice park.
ReplyDeletei used the first equation, too. for 4a, that is. i didn't get an old answer for 4a, but for 4b, i used the third equation (which is the only equation you can use out of the three), i got a negative velocity which doesn't make sense.
This comment has been removed by the author.
ReplyDeleteannie i just looked @ the problem and i got a negative and positive answer, same number different signs. easily its the positive answer. idk what you did wrong. i used equation #2 which worked better for me since you know everything in that eqn except the velocity.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteohhhhhhhhhhhhhhh, okay.
ReplyDeletei used the third equation.
thanks thanks thanks.
no problemo
ReplyDeletehey, john.
ReplyDeleteif you did #5b, tell me if i did it right.
i set up an equation for the stuntwoman by using V=Vo + at. then, i set up an equation for the horse by using d = Vot + 1/2at^2.
then i made them equal to each other and found t.
actually, nevermind.
ReplyDeletewhy did i do that.............
hmm for #5b its fairly simple. i did d=vot + 1/2at^2 because she falls there is no inital velocity so it ends up turning into d = 1/2 at^2
ReplyDeleteyou know every variable on that chart except time so its pretty easy, no need for 2 equations. also dont forget she is falling DOWN!
hahah, thank you !
ReplyDeleteCan anyone who's still awake help me with 5a?
ReplyDeleteomggggggggggggggg, i'm awake.
ReplyDeletewhat i did was solve 5b first.
and then use d=vt to find the distance.
Hey.. for #3 on the practice problem 2,
ReplyDeletewhich 2 equations should be used?
Okay, I thought that too. And yes, we're the nocturnals haha.
ReplyDeleteI didn't quite understand what #3 was asking.
ReplyDeleteWho's supposed to win ? - -
because it says 'largest lead the team B runnner can have...' and 'in order that team A runner not lose the race'
I figured out how to do #15, but everyone's probably sleeping.............
ReplyDelete