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.
I simply found average velocity of each runners, and then multiplied runner A's avg. velocity with runner B's 100m time. After that, you can see how much further runner A runs when he runs for same amount of time as runner B's 100m time
The time they're at the same height is equal to 15 meters (the length of the building) Since one ball is thrown upward from the ground and the other ball is dropped from a building, they meet in the middle. Therefore, you would use the d=(v intial)(t)+ 1/2(a)(t)^2 formula. You plug the info for the first ball in, then add it to the formula with info from the second ball plugged in and set that side equal to 15 m. After this, just type it into your calc and you should be good. =]
For #3. It reminds me of the lake problem we did in groups. they start at different ends (the ground and a building) and meet somewhere in the middle. A similar formula will work.
When the acceleration is unknown in a problem that involves an object going up or down, is the acceleration always -9.8m/s or since the negative only means direction would it be 9.8m/s if the object was thrown upwards?
Can anyone explain what equation to use for 14 on the MC? Im confused about that one.
Umm on #5 on the practice problems worksheet are you just trying to find the time it takes her to fall then use that to find the distance the horse runs in that time?
JWebs06 "When the acceleration is unknown in a problem that involves an object going up or down, is the acceleration always -9.8m/s or since the negative only means direction would it be 9.8m/s if the object was thrown upwards?"
Jake, I used the V^2=(initial velocity)^2+(2)(a)(d) for #2a. You first have to solve for V using this formula and the given initial velocity, then use the ending velocity you find in that equation as the new initial velocity. You use the same equation to find a distance. Then you add that distance to the distance given.
Lily, You have to set the two cars equal to each other using the d=Vot+1/2at^2 formula. Make sure you use t for the first car and t-6 for the second car.
Jenn, for #14, use the V=Vo^2+2ad equation. Think of it this way, the first ball is V (plug into Vo) and the second is 2V (also plug into Vo). You can see the difference in the heights by trying to do this.
JWebs,for #14 on the MC, I used the equation, v^2=vo^2 + 2ad and then rearranged the equation so that I was solving for d. Your final velocity for both will be 0 m/s.
For #3, I know you have to get the velocity of the falling ball first, but how do you do that without knowing the time or the distance? And would the initial velocity be 0 and we're trying the find the final..?
Asha, if you use the equation v^2=vo^2 + 2ad and solve the equation for d you end up with d=(v^2 - vo^2)/2a. Plug the two velocities into two seperate equations as the initial veolcities and use gravity for the acceleration. You should end up with a 4:1 ratio.
Does anybody get a negative velocity for 4b? I'm really doubting this answer because you can't have a negative velocity as an answer in this particular problem.
Oopsies. I was checking my work and found that I was wrong when I answered #1 on the problems. #1a. use the first formula. b.first formula again 3. just look for the velocity.
Which equation would you use for #2 on the multiple choice worksheet?
ReplyDelete/Kevin
ReplyDeleteI simply found average velocity of each runners, and then multiplied runner A's avg. velocity with runner B's 100m time. After that, you can see how much further runner A runs when he runs for same amount of time as runner B's 100m time
I used
ReplyDeleted=vt
because they are both speeders.
I use it to figure out how far runnerB would go in the time it took runnerA to finish.
What would the equation be in number 3 in the Practice Problems 2 worksheet?
ReplyDeleteWhat equation would you use in problem 1? I think I may have an answer but I am not sure so could someone just verify which they used for me?
ReplyDeleteDear Reema,
ReplyDeleteThe time they're at the same height is equal to 15 meters (the length of the building) Since one ball is thrown upward from the ground and the other ball is dropped from a building, they meet in the middle. Therefore, you would use the d=(v intial)(t)+ 1/2(a)(t)^2 formula. You plug the info for the first ball in, then add it to the formula with info from the second ball plugged in and set that side equal to 15 m. After this, just type it into your calc and you should be good. =]
I am soooo lost on problem 3 don't even know where to start maybe a formula could help.
ReplyDeleteFor equation #1
ReplyDeletea. I used t=d/v
b. v=d/t
I'm not sure for #1c. I keep getting really big numbers.
What equation should I use for #7 on the multiple choice worksheet?
ReplyDeletetyler-
ReplyDeletefor #1 part a, i used Vot + 1/2at^2.
and for part c of #1, i used V=Vo + at
ReplyDeleteFor #3.
ReplyDeleteIt reminds me of the lake problem we did in groups. they start at different ends (the ground and a building) and meet somewhere in the middle. A similar formula will work.
For #3
ReplyDeletesorry, I forgot to mention. calculate the velocity of the falling ball. and then plug it into a similar equation as the lake problem.
When the acceleration is unknown in a problem that involves an object going up or down, is the acceleration always -9.8m/s or since the negative only means direction would it be 9.8m/s if the object was thrown upwards?
ReplyDeleteCan anyone explain what equation to use for 14 on the MC? Im confused about that one.
This comment has been removed by the author.
ReplyDeleteUmm on #5 on the practice problems worksheet are you just trying to find the time it takes her to fall then use that to find the distance the horse runs in that time?
ReplyDeleteOn the practice problems 2 sheet... My answer to 2a is really an awkward number. I used this equation, d=(v intial)(t)+ 1/2(a)(t)^2, what went wrong?
ReplyDeleteJWebs06
ReplyDelete"When the acceleration is unknown in a problem that involves an object going up or down, is the acceleration always -9.8m/s or since the negative only means direction would it be 9.8m/s if the object was thrown upwards?"
Gravity always works downward.
Jake,
ReplyDeleteI used the V^2=(initial velocity)^2+(2)(a)(d) for #2a. You first have to solve for V using this formula and the given initial velocity, then use the ending velocity you find in that equation as the new initial velocity. You use the same equation to find a distance. Then you add that distance to the distance given.
Lily,
ReplyDeleteYou have to set the two cars equal to each other using the d=Vot+1/2at^2 formula. Make sure you use t for the first car and t-6 for the second car.
Megan, yes, thats what i did.
ReplyDeleteEunice, I used the first equation too. Make sure your plugging in everything right. Maybe you forgot to square the t^2?
ReplyDeleteJenn, for #14, use the V=Vo^2+2ad equation. Think of it this way, the first ball is V (plug into Vo) and the second is 2V (also plug into Vo). You can see the difference in the heights by trying to do this.
ReplyDeleteJWebs,for #14 on the MC, I used the equation, v^2=vo^2 + 2ad and then rearranged the equation so that I was solving for d. Your final velocity for both will be 0 m/s.
ReplyDeleteFor #3, I know you have to get the velocity of the falling ball first, but how do you do that without knowing the time or the distance? And would the initial velocity be 0 and we're trying the find the final..?
ReplyDeletecan someone please explain 4b on the practice problems? What eqn do you use?
ReplyDeleteDid anyone get #14 in the m.c. problems? I keep trying different equations, but none of them make any sense. I don't get why the ratio is 4:1
ReplyDeleteAsha, if you use the equation v^2=vo^2 + 2ad and solve the equation for d you end up with d=(v^2 - vo^2)/2a. Plug the two velocities into two seperate equations as the initial veolcities and use gravity for the acceleration. You should end up with a 4:1 ratio.
ReplyDeleteJackie,
ReplyDeleteI used V=Vo+at. You have to use whatever your answer was for 4a.
Does anybody get a negative velocity for 4b? I'm really doubting this answer because you can't have a negative velocity as an answer in this particular problem.
ReplyDeletedid anyone get number five on practice problems2? what equation do we use?
ReplyDeleteOopsies. I was checking my work and found that I was wrong when I answered #1 on the problems.
ReplyDelete#1a. use the first formula.
b.first formula again
3. just look for the velocity.
Sorry. my bad.
/Dan
ReplyDeleteFor 4b, I used the formula V^2=Vo^2+2ad. In Part A, you should've gotten the Vo, so just plug and chug from there. :)
-Christine Q
what do you use for t or d after engine stops in 2a?
ReplyDeletewhy is the distance always 4x longer when the acceleration or speed is twice as fast?
ReplyDeleteshouldnt it just be 2x
like problems 6,9,14?