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Old Sep 21st 2009, 08:35 AM   #1
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Problems that I need help with

Okay, so we have some problems that I need to submit, ASAP! So the first one says:

Over a period of 11.0 seconds a plane changes its velocity from 35.6 m/s to 22.1 m/s. How far does the plane go during this time interval?

Now my first attempt was to just subtract the velocities and multiply that by the time but I got that wrong and also someone told me that you need to use one of the kinematic equations ( v=at +vi ; x =1/2at^2+vit+xi ; vf^2 + 2ax +vi )but i dont know which to use or what to plug in to which ones.


Another problem I have is:

A cart moving at 15.7 m/s begins to coast up a hill with a uniform acceleration of -2.7 m/s2. How far does it travel in 9.1 s?

My attempt was to ignore the first velocity and just multiply the acceleration by the time and that would leave with just meters which would be the answer but I got that wrong, so what sho9uld I do. ( I'm thinking it involves the use of the kinematic equations again - v=at +vi ; x =1/2at^2+vit+xi ; vf^2 + 2ax +vi )


My last problem is:

A cart moving at 19.8 m/s begins to coast up a hill with a uniform acceleration of -1.5 m/s2. How fast is it moving after it has traveled a total distance of 30.8 m?

Now with this one, I honestly don't know where to begin.
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I know this is kind of a lot but if you can help me with these, even if its just one or two of them, please do even though I really need help with all. Any assistance would be greatly appreciated. =D Thanks.
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Old Sep 21st 2009, 10:40 AM   #2
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Over a period of 11.0 seconds a plane changes its velocity from 35.6 m/s to 22.1 m/s. How far does the plane go during this time interval?

s= displacement
t=time
u=initial velocity
v=final velocity

s= (u+v)/2 x t

Answer = 317.35 m



With regard the cart questions have you got the acceleration figures right? you mention acceleration but show minus figures?
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Old Sep 21st 2009, 11:21 AM   #3
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Originally Posted by Paul46 View Post
Over a period of 11.0 seconds a plane changes its velocity from 35.6 m/s to 22.1 m/s. How far does the plane go during this time interval?

s= displacement
t=time
u=initial velocity
v=final velocity

s= (u+v)/2 x t

Answer = 317.35 m



With regard the cart questions have you got the acceleration figures right? you mention acceleration but show minus figures?
yeah i was wondering about that too, but that's the way the problem was given to me, negatively.

and thank you for answering that question, i appreciate it. does that formula always work? because we've never been introduced to that one.

Last edited by INeedHelp; Sep 21st 2009 at 11:36 AM.
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Old Sep 21st 2009, 02:03 PM   #4
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yeah the formula always works if there is a change in velocity.

regarding the other questions, i can only take it they are negative if you say so.


A cart moving at 15.7 m/s begins to coast up a hill with a uniform acceleration of -2.7 m/s2. How far does it travel in 9.1 s?


s= displacement
t=time
u=initial velocity
v=final velocity
a=acceleration

s= ut + (1/2)at^2

s= 15.7 x 9.1 + (1/2) x -2.7 x 9.1^2 = 31.0765 m


A cart moving at 19.8 m/s begins to coast up a hill with a uniform acceleration of -1.5 m/s2. How fast is it moving after it has traveled a total distance of 30.8 m?


v^2 = u^2 + 2as

392.04 + 2 x -1.5 x 30.8 = 299.64 sq-rt = 17.31 m/s
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Old Sep 22nd 2009, 05:57 AM   #5
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Originally Posted by Paul46 View Post
yeah the formula always works if there is a change in velocity.

regarding the other questions, i can only take it they are negative if you say so.


A cart moving at 15.7 m/s begins to coast up a hill with a uniform acceleration of -2.7 m/s2. How far does it travel in 9.1 s?


s= displacement
t=time
u=initial velocity
v=final velocity
a=acceleration

s= ut + (1/2)at^2

s= 15.7 x 9.1 + (1/2) x -2.7 x 9.1^2 = 31.0765 m


A cart moving at 19.8 m/s begins to coast up a hill with a uniform acceleration of -1.5 m/s2. How fast is it moving after it has traveled a total distance of 30.8 m?


v^2 = u^2 + 2as

392.04 + 2 x -1.5 x 30.8 = 299.64 sq-rt = 17.31 m/s
thank you! i appreciate the help.
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