Physics, 26.03.2020 01:56 Thiskid100

The 10-mg barge b supports a 2-mg automobile

a. if someone drives the automobile to the other side of the barge, determine how far the barge moves. neglect the resistance of the water.

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Answer from: ayindejoy

The image showing the barge is missing, so i have attached it.

It moves 6.667m in the right direction.

Explanation:

To solve this, we will assume that the barge will move to the left.

If we apply conservation of linear momentum to this question, we'll arrive at ;

MaVa + MbVb = MaVa' + MbVb'

Where;

Ma = mass of automobile

Mb = mass of barge

Va = Initial velocity of automobile

Vb = Initial velocity of barge

Va' = Final velocity of automobile

Vb' = Final velocity of barge

Now;

Mb = 10Mg = 10 x 10³g = 10,000 g

Ma = 2 Mg = 2 x 10³ = 2,000g

Va' = 0 m/s

Vb' = 0 m/s

Thus;

2,000Va + 10,000Vb = 0

2,000Va = - 10,000Vb

Va = - 10,000Vb/2,000

Va = - 5Vb

Now, velocity is distance/time and by experience, with same time, we can also say that;

Sa = -5Sb

From the image i attached, the relative distance between A and B is 40m

Thus, Sa - Sb = 40

From earlierSa = - 5Sb

Thus,

-5Sb - Sb = 40

-6 Sb = 40

Sb = -40/6 = - 6.667m

Since, this is negative it means it is in an opposite direction to the left which we chose.

Thus,

Sb moves 6.667m in the right direction.

The 10-mg barge b supports a 2-mg automobile a. if someone drives the automobile to the other side o

Answer from: Quest

Iwas going to ask that quesytion sorry dude i dont know : ( : /

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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The 10-mg barge b supports a 2-mg automobile

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