A jetliner leaves san francisco for new york, 4600 km away. with a strong tailwind, its speed is 1100 km/hr. at the same time, a
second jet leaves new york for san francisco. flying into the wind, it makes only 700 km/hr. when and where do the two planes pass?
1 answer:
First, make an illustration of the problem. In reality, the direction traveled by the planes is not linear. Strictly speaking, it should curve out following the shape of the sphere. But I suppose, we could do away with that because we are given with the quantity for distances. I'm just showing a guide to see where the planes would meet.
So, we know that the total distance is 4,600 km. It would consequently mean that x and y would equal to 4,600. The distances x and y could be determined by multiplying their respective speeds to the time. Since they meet at the same time, let's denote it as t.
1,100t + 700t = 4,600
Solving for t,
1,800t = 4600
t = 4600/1800
t = 2.556 hours
So, we know that the two planes would meet after 2.56 hours from the time they leave the airports. Next, we have to know where they meet exactly.
1,100(2.556) = 2,812 km
700(2.556) = 1,788 km
Therefore, they meet at approximately 2,812 km from San Francisco airport, or 1,788 km from New York airport after 2.56 hours since they left their respective ports.
So, we know that the total distance is 4,600 km. It would consequently mean that x and y would equal to 4,600. The distances x and y could be determined by multiplying their respective speeds to the time. Since they meet at the same time, let's denote it as t.
1,100t + 700t = 4,600
Solving for t,
1,800t = 4600
t = 4600/1800
t = 2.556 hours
So, we know that the two planes would meet after 2.56 hours from the time they leave the airports. Next, we have to know where they meet exactly.
1,100(2.556) = 2,812 km
700(2.556) = 1,788 km
Therefore, they meet at approximately 2,812 km from San Francisco airport, or 1,788 km from New York airport after 2.56 hours since they left their respective ports.
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Explanation:
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And Final velocity at that point be v
Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,
Let '' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,
At that insitant,
- initial velocity of the rocket = v = 195.5 m/s.
- a =
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From the kinematics,
Hence the maximum altitude attained by the rocket from the ground is
Answer:
,
Explanation:
First we build the Free Body Diagram (please see first image for further details) associated with the mass, we notice that system consist of a three forces that form a right triangle (please see second image for further details): (i) The weight of the mass, (ii) two tensions.
The requested tension and angle can be found by the following trigonometrical and geometrical expressions:
(1)
(2)
Where:
- Weight of the mass, measured in newtons.
,
- Tensions from the mass, measured in newtons.
If we know that and
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