EK 1001 Physics #58

[MedPR]

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I couldn't find another thread about this problem, so here goes.

This is on page 6 for those of you with EK 1001 Physics.

A particle moving at 5m/s reverses its direction in 1s to move at 5m/s in the opposite direction. If its acceleration is constant, what is its displacement at 0.5 seconds?

A: 0m
B: 1.25m
C: 2.5m
D: 5m

I got 2.5m because intuitively that's what I come up with. According to the answer, 2.5 is wrong, and 1.25 is correct. Here is the solution explanation.

UUse x=vot+1/2at^2, where a=(vfinal-vinitial)/t, but use 1/2 for t in the first equation because we want the speed halfway through the trip.

When I do that, I get x=5(.5)+ [(5-5)/2]t, which is x=2.5 +0t which is x=2.5..

Edit: Just realized the sign on my vfinal was wrong, but it still comes out to x=0

x=5(.5) + [(-5-5)/2]t = 2.5 - 5*.5 = 0

 

[MrNeuro]

I couldn't find another thread about this problem, so here goes.

This is on page 6 for those of you with EK 1001 Physics.

A particle moving at 5m/s reverses its direction in 1s to move at 5m/s in the opposite direction. If its acceleration is constant, what is its displacement at 0.5 seconds?

A: 0m
B: 1.25m
C: 2.5m
D: 5m

I got 2.5m because intuitively that's what I come up with. According to the answer, 2.5 is wrong, and 1.25 is correct. Here is the solution explanation.

UUse x=vot+1/2at^2, where a=(vfinal-vinitial)/t, but use 1/2 for t in the first equation because we want the speed halfway through the trip.

When I do that, I get x=5(.5)+ [(5-5)/2]t, which is x=2.5 +0t which is x=2.5..

Edit: Just realized the sign on my vfinal was wrong, but it still comes out to x=0

x=5(.5) + [(-5-5)/2]t = 2.5 - 5*.5 = 0

use this instead
s=1/2(u+v)t

 

[MedPR]

use this instead
s=1/2(u+v)t

Not sure what that equation is. Same as x=vavg * t?

If yes, s = 1/2(5-5)(.5) = 0.

Or, if you don't give vf a negative (it's in the opposite direction, so one of the velocities must be positive, one must be negative) you get 1/2(5+5)(.5) = .5*10*.5 = 5*.5 = 2.5. Still don't get the correct answer of 1.25.

 

[chiddler]

Not sure what that equation is. Same as x=vavg * t?

If yes, s = 1/2(5-5)(.5) = 0.

Or, if you don't give vf a negative (it's in the opposite direction, so one of the velocities must be positive, one must be negative) you get 1/2(5+5)(.5) = .5*10*.5 = 5*.5 = 2.5. Still don't get the correct answer of 1.25.

No dude it's one of the kinematic equations (that you should know!).

I'm not sure how to answer your question because all the kinematic equations that have both Vf and Vi in them would be equal to 0 if Vi = -Vf as we see with this problem.

I'm not confident in my knowledge enough yet to assure you, but maybe the book is wrong?

 

[ljc]

My method:

5m/s to -5m/s in 1.0 seconds = -10m/s^2 acceleration. Given that are looking at 0.5s, we just need to multiply -10 by 0.5, and we get -5. This means at 0.5s we are going 0m/s. This is pretty obvious, though let me know if you need more explanation here.

Next, recognize that initial velocity is 5, final velocity is 0, giving an average velocity of 2.5 over our time frame. Now, since our average velocity is 2.5, and we are travelling for 0.5s, we multiply them together: 2.5m/s*0.5s=1.25m.

Hope that helps.

Edit: and to answer the above poster - Vi is not equal to -Vf, because we are looking at the time frame of 0s-0.5s. If our time frame were 0s-1s, Vi would be equal to -Vf. In this case, though, we define Vf to be at 0.5s, so it ends up being 0m/s, and Vi is +5m/s.

 

[chiddler]

My method:

5m/s to -5m/s in 1.0 seconds = -10m/s^2 acceleration. Given that are looking at 0.5s, we just need to multiply -10 by 0.5, and we get -5. This means at 0.5s we are going 0m/s. This is pretty obvious, though let me know if you need more explanation here.

Next, recognize that initial velocity is 5, final velocity is 0, giving an average velocity of 2.5 over our time frame. Now, since our average velocity is 2.5, and we are travelling for 0.5s, we multiply them together: 2.5m/s*0.5s=1.25m.

Hope that helps.

Edit: and to answer the above poster - Vi is not equal to -Vf, because we are looking at the time frame of 0s-0.5s. If our time frame were 0s-1s, Vi would be equal to -Vf. In this case, though, we define Vf to be at 0.5s, so it ends up being 0m/s, and Vi is +5m/s.

good point i forgot that.

 

[MedPR]

My method:

5m/s to -5m/s in 1.0 seconds = -10m/s^2 acceleration. Given that are looking at 0.5s, we just need to multiply -10 by 0.5, and we get -5. This means at 0.5s we are going 0m/s. This is pretty obvious, though let me know if you need more explanation here.

Next, recognize that initial velocity is 5, final velocity is 0, giving an average velocity of 2.5 over our time frame. Now, since our average velocity is 2.5, and we are travelling for 0.5s, we multiply them together: 2.5m/s*0.5s=1.25m.

Hope that helps.

Edit: and to answer the above poster - Vi is not equal to -Vf, because we are looking at the time frame of 0s-0.5s. If our time frame were 0s-1s, Vi would be equal to -Vf. In this case, though, we define Vf to be at 0.5s, so it ends up being 0m/s, and Vi is +5m/s.

I don't get how you can set the final velocity = to 0.

Edit: Nevermind, I got it. I completely overlooked the 5m/s to -5m/s in 1 second == 0m/s at .5seconds from the problem.

Thank you!

 

[MedPR]

No dude it's one of the kinematic equations (that you should know!).

I'm not sure how to answer your question because all the kinematic equations that have both Vf and Vi in them would be equal to 0 if Vi = -Vf as we see with this problem.

I'm not confident in my knowledge enough yet to assure you, but maybe the book is wrong?

Are you serious? x=vavgt is a kinematics equation just in a different form (different letters) than he wrote and I wasn't sure if he meant the same equation.

 

[study pray love]

in order for an object to reverse directions its velocity must first be reduced to zero.

think of driving your car: if you are moving in "drive" North and want to go South you must first apply your breaks in order to reverse gears and move South.

therefore to find the displacement for half the time:
1/2time=.5 sec
Vo= 5 m/s
Vf=0 (for the above reason)

*****Vavg=d/t*****

(Vo+Vf)/2= d/t
so
(5+0)/2=d/t
so
2.25=d/.5 and THEREFORE: d=1.25 for half the time

 

[grapepopsicle]

Another way to look at it:

d = vot + 1/2at^2

vo = 5
a = (vf - vi)/t = (-5 - 5)/1 sec = -10m/s^2

d = 5(.5) + (1/2)(-10)(0.5^2)
= 2.5 + (1/2)(-10)(.25)
= 2.5 + (1/2)(-2.5)
= 2.5 - 1.25
= 1.25 m