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Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other ?
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Answers
Given :-
Three particles A, B and C are situated at the vertices of an equilateral triangle ABC.The side of the triangle is d .A always has its velocity along AB, B along BC and C along CA.To Find :-
Time at which the particles will meet .Solution :-
See attachment .
Firstly break the velocity of A , into it's vertical components . That will be ,
[tex] v_A = v_A cos60^o + v_A sin60^o [/tex]
Now the component , [tex]v_A cos60^o [/tex] coincides on side AC and is directed from A to C . Along CA, the velocity of particle C is directed . So both are in opposite directions.
So , the relative velocity of C with respect to A will be ,
[tex] v_{CA}= v_C - v_A [/tex]
Substitute ,
[tex] v_{CA}= v - ( - v cos 60^o) [/tex]
Value of cos 60° is 1/2 ,
[tex] v_{CA}= v + v\bigg(\dfrac{1}{2}\bigg)=v +\dfrac{v}{2}[/tex]
Add the terms,
[tex] v_{CA}= \dfrac{2v + v}{2}=\dfrac{3v}{2}[/tex]
Hence , now we can find the time taken by A and C ,to meet each other will be ,
Since the speed is constant , we can use distance = speed*time , as ;[tex]s = v t [/tex]
Substitute ,
[tex]d = \dfrac{3v}{2} t [/tex]
Multiplying both sides by 2 ,
[tex] 2d = 3vt [/tex]
Dividing both sides by 3v ,
[tex] t = \dfrac{2d}{3v} [/tex]
Similarly , taking any two particles you will get the same time .
Hence the particles meet after 2d/3v time .
I hope this helps .
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/Z5g8PLR1Ud9OrJslRW96o3k0rFIbOsT0.jpeg)
Answer:
never
Explanation:
Particles moving on paths that don't meet at a point will be particles that never meet. Even if the paths meet at a point, which these don't, the particles may not meet if they arrive at different times.
__
Consider the diagrams in the attachments. In order, they show the positions of the particles shortly after takeoff, midway along the line, and after they have gone farther than the length of the line. The point of closest approach is shown in the second attachment, when the particles are halfway along the line.
Consider particles A and C. Their north/south direction is 'north' for each of them. They both have the same velocity in the 'north' direction, but have started at different points, so can never meet. Those two particles have opposite velocities in the east/west direction, so can pass each other (which they do in the second attachment), but cannot meet.
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/zTC2NCzcFLmVstKZDxpzyeqOt2E09c5i.png)
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/7ZocH6SVq9vsgplLVSFQAO6Uv1BN5mH7.png)
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/Jb0lRj55AhWznP42LcU76e4pvAFx8niG.png)
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a) The expression to determine the velocity of cart A is [tex]v_{A} = 2\cdot \sqrt{g\cdot H}[/tex].
b) The expression to determine the velocity of cart B is
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[tex]v_{A} = 2\cdot \sqrt{g\cdot H}[/tex]
The expression to determine the velocity of cart A is [tex]v_{A} = 2\cdot \sqrt{g\cdot H}[/tex]. [tex]\blacksquare[/tex]
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The expression to determine the velocity of cart B is
[tex]v_{B} =\sqrt{2\cdot g\cdot H }[/tex]. [tex]\blacksquare[/tex]
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Answer:
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Answers
Answer:
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Answers
Answer:
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Answer:
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Answers
Answer:
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Explanation:
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Answer:
True
Explanation:
Answer:
The answer is T (True)
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Answer:
1/3 kg
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Answer:
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Answers
Answer:
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Explanation:
Answer:
No.
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Answers
Answer:
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Motion of Car
Speed
Distance
0 meters
50 meters
100 meters
150 meters
meters
250 meters
Time
0.0 seconds
4.3 seconds
8.6 seconds
12.9 seconds
17.2 seconds
21.5 seconds
200
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B. The car was travelling at a constant speed.
C. The car was slowing down.
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PLEASE ANSWER ILL GIVE BRAINLIEST !!!!!
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what is this math i do not get it pls helppppp
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Answer:
where is the problem u need help with
Explanation:
(: pls mark brainliest
Answer: math is an area of knowledge that includes the study of such topics. as numbers and formulas related structures, algebra and geometry also calculus and analysis
Explanation:
The Big Bang theory states that _____.
the universe was created from a cosmic explosion
the universe was formed from a collision between two galaxies
the universe is expanding and will continue to expand
the sun will eventually burn out and collapse
Answers
Answer:
A
Explanation:
The universe was created from a cosmic explosion.
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Answers
Answer:
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Explanation:
Answer:
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1. A rock of granite has a mass of 50 kg. if it’s weight in water
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Answers
Answer:
The first part of the question is asking about BUOYANT FORCE or UPTHRUST.
Upthrust =TRUE WEIGHT-APPARENT WEIGHT
TRUE WEIGHT=mg
TRUE weight=50kg×10m/s²
=500N
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please mark brainliest, hope it helped
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Answers
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pig-footed bandicoot
humans
eremophila plant
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Answers
Answer:
Kangaroo should be the answer.
Answer:
Kangaroo So there's your answer
Explanation:
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calculate the GPE of a rock with a mass of 55kg on a cliff that is 27m high
Answers
Answer:
P=mgh
P=mghm=55
P=mghm=55g=9.8 or ~10
P=mghm=55g=9.8 or ~10h=27
Explanation:
hope it will help you have a great day bye and Mark brainlist if the answer is correct
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