Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
First segment: The airplane is descending slowly, with time and distance at a 3:1 ratio. It is probably approaching the airport
Second segment: The airplane is maintaining its altitude. It is probably waiting for the other planes to clear away
Third segment:The airplane descending rapidly, with time and distance at a 1:2 ratio. It is probably landing
Sorry, my bad.
To find acceleration, we use v2-v1 /t
40mph - 20mph / 10 min
20mph/10min
We can convert this to 20/6 miles per 10 minutes, and cancel out the 10 min to 20/6 miles, which is 10/3 miles
Second segment: The airplane is maintaining its altitude. It is probably waiting for the other planes to clear away
Third segment:The airplane descending rapidly, with time and distance at a 1:2 ratio. It is probably landing
Sorry, my bad.
To find acceleration, we use v2-v1 /t
40mph - 20mph / 10 min
20mph/10min
We can convert this to 20/6 miles per 10 minutes, and cancel out the 10 min to 20/6 miles, which is 10/3 miles
Acceleration = (change in speed) / (time for the change)
(Note: That's all the Physics there is to this problem.
The rest is all arithmetic.)
=======================================
change in speed = (40 - 20) miles/hour = -20 miles/hour
time for the change = 10 minutes
Acceleration = (-20 miles/hour) / (10 minutes) =
-2 miles/hour per minute .
That's a perfectly good and technically correct expression for acceleration.
But obviously the units might make some people dizzy. So let's try to
clean it up a little.
Notice that 10 minutes is 1/6 of an hour.
So we could write the acceleration as
Acceleration = (-20 miles/hour) / (1/6 hour) = -120 miles/hour per hour =
-120 miles/hour² .
You could convert this into any units you like. It's really not a physics problem
any more, it's just an exercise in converting units.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.