Physics -- Check this please

Hey,

I have this problem in physics which I know the answer to (answers are in the back), but I'm getting a different answer. At this point I'm thinking the back of the book is wrong ;)

The question:

Object A is dropped by an Object B that is rising steadily at .5 m/s.

a) After 2.5 s, what is the velocity of Object A?
b) How far below Object B is Object A after 2.5 s?


solution to a:

vf = vi + at
vf = .5 + (-9.8 m/s)(2.5 s)
vf = -24 m/s (Correct answer)


solution to b:

x = (.5 m/s)(2.5 s) + .5(-9.8 m/s)(2.5 s)^2
x = -29.375 (wrong answer)

The correct answer should be 31 m, but how in the world did they get that? I've tried many of the formulas I have and still come up with the same answer.


Can anyone shed some light on part B?

Yeah, you're answer is correct. I'm guessing they just forgot to account for Object B moving. I used the 1/2*a*t^2 formula for the distance of object A and got 30.625m, rounded to 31m. So I guess thats what they did.

Just a question, does it give any other info about the objects like whether they started out at the same height or something? I see that question as very unclear lol.

Quote:

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Originally Posted by System_Error

Hey,

I have this problem in physics which I know the answer to (answers are in the back), but I'm getting a different answer. At this point I'm thinking the back of the book is wrong ;)

The question:

Object A is dropped by an Object B that is rising steadily at .5 m/s.

a) After 2.5 s, what is the velocity of Object A?
b) How far below Object B is Object A after 2.5 s?


solution to a:

vf = vi + at
vf = .5 + (-9.8 m/s)(2.5 s)
vf = -24 m/s (Correct answer)


solution to b:

x = (.5 m/s)(2.5 s) + .5(-9.8 m/s)(2.5 s)^2
x = -29.375 (wrong answer)

The correct answer should be 31 m, but how in the world did they get that? I've tried many of the formulas I have and still come up with the same answer.


Can anyone shed some light on part B?

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I think 31m is the correct answer,

Because

(1) Object B is moving upwards and that same speed (0.5m/s) has to A

(2) If A drop from B, first B goes to upwards until its speed become zero.

V^2 = U^2 + 2*F*S
V - Last speed = 0
U - Start speed = 0.5m/s
F - Acceleration = -9.8 ( - sign used because I used the equation to upwards )

So S1 = 0.01m
Actually you can neglect that,

Then, it's moved down, with acceleration isn't it?

For B,

S2 = 0.5 * 2.5 = 1.25m

For A,

Using v = u + f*t find v, at that time u = zero, because A is at the top point, isn't it?

So, v = 24.5m/s

It travels

v^2 = u^2 + 2*f*s

s = 30.625

So the distance between A and B at 2.5s is = S2 + s
= 1.25 + 30.625
= 31.875 m

If you think that S1 distance then the answer is almost 31m, because at 2.5s B goes upwards and then starts to moved down.

Check this friend; I'll try to check completely later for you.

I think this is ok. If not please let me now. :wave:

My answer is also 30.625 m. I don't know if it's correct, but if they state the correct answer is 31 m I don't think they've rounded up, it would be misleading, so maybe we've all overlooked something.

At any rate the way to solve the problem seems quite straightforward. If you call y the vertical axis and take as positive the upward direction, then the required distance is

D = y_B(2.5 s) - y_A(2.5 s)

and applying the basic equations, assuming y=0 and t=0 when object A is released by object B:

y_B(2.5 s) = y_B0 + v_B0t + 1/2 a_Bt²
y_A(2.5 s) = y_A0 + v_A0t + 1/2 a_At²

Now,

y_B0= 0
y_A0= 0
v_B0= 0.5 m/s
v_A0= 0.5 m/s
a_B=0
a_A=-9.8 m/s²
t = 2.5 s

and substituting you get:

y_B= 1.25 m
y_A= -29.375 m
so that D = 1.25 m - (-29.375 m) = 30.675 m

Thanks guys. It seems like the back of the book gave the wrong answer. I ask him about the problem and he worked it out. He got the exact same answer as we did ;)


-29.375 and then you add (subtract) the 1.25 meters from Object A.

Quote:

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Originally Posted by System_Error

Thanks guys. It seems like the back of the book gave the wrong answer...

-29.375 and then you add (subtract) the 1.25 meters from Object A.

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???

Then surely the book gave the right answer of 29.375 + 1.25 = 30.625 meters, rounded to 31...?

The book does, indeed, give the right answer.

Quote:

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Originally Posted by Dross

???

Then surely the book gave the right answer of 29.375 + 1.25 = 30.625 meters, rounded to 31...?

The book does, indeed, give the right answer.

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Why would it round the answer? It doesn't on any other problem?