How do you calculate the effects of crosswinds on a plane's velocity?

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To calculate the effects of crosswinds on a plane's velocity, the plane's southward speed of 100 m/s and a westward crosswind of 25 m/s combine vectorially, resulting in a total velocity of approximately 103.08 m/s at an angle of 14 degrees south of west. To maintain a southward velocity of 100 m/s, the pilot must adjust the heading accordingly. In a separate problem, a car traveling at 80 km/h is pursued by a policeman who accelerates to 100 km/h. After 15 seconds, the distance between them can be calculated, and using the relative speed of 20 km/h, the time for the policeman to catch up can be determined with the formula t = X/20. Understanding these principles of motion and relative velocity is essential for solving such problems effectively.
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I am upto Motion in my studies, but the last time i studied it was a year ago and i am trying to get into swing of it again.

2 question that bug me at the moment.

1.
A plane is flying due south at 100m/s on an initially still day. Then a crosswind with a velocity of 25m/s towards the west beings to blow.
- What is the velocity of the plane under the influence of this crosswind?
- In which direction should the pilot steer the plane to maintain a velovity of 100m/s south?

I used a simple vector sum:
vplane = 100
vair = 25
vsum = square root((vplane^2)+(vair^2))
i got vsum = 103.08m/s

and the direction is South 14Degrees West.

I think that's right, the answer states its 10^3 ..but i think they accidentely put the 3 as a power. 10^3 not 103

2.
A car traveling with a constant speed of 80 km/h passes a stationary motercycle policeman. The policeman sets off in pursuit, acceserating to 80km/h in 10s and reaching a contstant speed of 100km/h after a further 5s. At what time will the policeman catchup with the car?

This question really made me think of last years studies, but i couldn't get my head around it. After the 15 seconds i can find out how far the car and policemen have traveled, but can't figure it out further.

Thanks for your time,
Cummings
Cummings@softhome.net
 
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I will go for question 2.
First of all, i will assume that the acceleration in the first 10 seconds is constant, and that it is also constant (but not with the same value) in the next 5 seconds.
You said you can figure out how far the policeman and car went after the first 15 seconds.
Now, (i will leave the numbers work for you) suppose the policeman and the car are X kilometers apart after the 15 seconds, the speed of the car will be constant (it didn't accelerate) and the speed of the policeman will also be constant (acceleration stopped after the first 15 seconds), all you have to do is to figure out the relative speed of policeman as seen from the car (which obviously is 100-80=20 km/h), it is like if we are assuming that the car is stationary, and the policeman is moving at the speed of 20 km/h, the distance between them is X as we said, so simply :
20=X/t
20t=X
t=X/20

It is this simple ! you did the hard work, now you only need to plug the values in that equation :smile:.
 
20kmh is the speed
we know the distance and thus can find the time.

Okies..easy done. Thanks :)
 
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