Physics -- Check this please

[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?

[fartman_900]

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.

[eranga262154]

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?

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.

[krtxmrtz]

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

[System_Error]

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.

[Dross]

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.

???

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.

[System_Error]

???

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.

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