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Carnegie Mellon University

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Problem 26 Hard Difficulty

In An airplane maintains a speed of 630 $\mathrm{km} / \mathrm{h}$ relative to the air it is flying through, as it makes a trip to a city 750 $\mathrm{km}$ away to the north. (a) What time interval is required for the trip if the plane flies through a headwind blowing at 35.0 $\mathrm{km} / \mathrm{h}$ toward the south? (b) What time interval is required if there is a tailwind with the same speed? (c) What time interval is required if there is a crosswind blowing at 35.0 $\mathrm{km} / \mathrm{h}$ the east relative to the ground?

Answer

(a) 1.26$\mathrm { h }$
(b) $t = 1.13 \mathrm { h }$
(c) $t = 1.19 \mathrm { h }$

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Video Transcript

The velocity of the plane relative to the air is going to be equaling 630 kilometres per hour. J hat when we can say the velocity of the headwind. Um, the air speed essentially is this scent is represented by Visa A. This would be equaling 35.0 kilometers per hour. Uh, negative J hat. And so we consider the velocity of the plane. When it hits, the headwind is gonna be equaling. Two, 596 130 minus 35 605 195 kilometers per hour. J hat. And we can say that. Then we know that the velocity of the plane when it hits the wind is gonna be equal in the distance about about time. And so the time is gonna be equaling 750 kilometers divided by 595 kilometers per hour. So it's gonna take the plane. Excuse me 1.26 hours to reach the city, so this would be your answer for party for part B. Then we can say the velocity of the plane on the way back is again equaling for 630 kilometres per hour. We're not on the way back, but this is this would be to the same trip. However, hear the wind is helping you. So, uh, this would be J hat and the the velocity of the air would be equaling 35 0.0 kilometers per hour. J hat. And so when it hits the when it hits the air, this is going to be 665 kilometers per hour. J hat That would be the speed of the airplane. And so now we can say that the time is gonna be equally 750 kilometers divided by 665 kilometers per hour. This is equally would 0.13 hours. So it takes a little bit less time to reach the city. If the tailwind is with you. And so we can say that then for part c ah, we could say that without, uh without any wind here, the velocity of the plane is equaling against 630 kilometers per hour. But here the velocity of the wind of the air rather the airspeed is equaling 35 0.0 kilometers per hour. However, this is I had this would be J hat. So these air in completely different directions and so the air is not gonna affect the velocity of the plane. And so the time would simply be equaling 750 kilometers per hour, divided by solely the velocity of the plane. We are not accounting for the wind because the wind is at a completely different direction. And so this would be divided by 630 kilometres per hour. This is supposed to be somewhere in between. So 1.19 hours, which makes sense. This is a little bit more than what we got for part B, but a little bit less than what we got for party. And so that would be a final answer. That is the end of the solution. Thank you for watching

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