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Old Jul 15th 2018, 06:47 AM   #1
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Homework question .......please help me

Two brothers, one with a mass of m and two with a mass of 2m,
* Each stands on a skateboard and pulls each other by force with a massless rope.
The floor is horizontal and the friction between the ceiling and the floor can be neglected.
At the moment t = 0 the distance between the brothers is L and their velocity.
Given that the acceleration of the younger brother is a.

Given:

L = 2 [m]

a = 0.9 [m / sec ^ 2]

At what time (in seconds) will the brothers clash?
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Old Jul 15th 2018, 08:00 AM   #2
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Originally Posted by fares View Post
Two brothers, one with a mass of m and two with a mass of 2m,
* Each stands on a skateboard and pulls each other by force with a massless rope.
The floor is horizontal and the friction between the ceiling and the floor can be neglected.
At the moment t = 0 the distance between the brothers is L and their velocity.
Given that the acceleration of the younger brother is a.

Given:

L = 2 [m]

a = 0.9 [m / sec ^ 2]

At what time (in seconds) will the brothers clash?
What have you been able to do with this so far?

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Old Jul 16th 2018, 04:58 AM   #3
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Originally Posted by fares View Post
Two brothers, one with a mass of m and two with a mass of 2m,
* Each stands on a skateboard and pulls each other by force with a massless rope.
The floor is horizontal and the friction between the ceiling and the floor can be neglected.
"Friction between the ceiling and the floor"? Why should the ceiling and the floor ever come into contact? Do you mean between the skateboards and the floor?

At the moment t = 0 the distance between the brother's velocity.
"And their velocity" is what?

Given that the acceleration of the younger brother is a.

Given:

L = 2 [m]

a = 0.9 [m / sec ^ 2]

At what time (in seconds) will the brothers clash?
a= 0.9 m/s^2 is the acceleration of each brother toward each other? Their mutual velocity is the anti-derivative of acceleration. And the distance between them is the anti-derivative of velocity. They will "class" (a peculiar word here. I would have said "crash" or "collide") when that distance is 0.
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