Question

A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. (a) What is the acceleration experienced by a pilot flying in a circle of constant radius at a constant speed of 485 m/s if the radius of the circle is 3470 m? (b) Does the pilot make the turn without blacking out?

Answer 1

@ganeshie8

Answer 2

@Shalante

Answer 3

Do you take (485m/s)^2/3470m and get 67.7881844m/s^2?

Answer 4

I think this might be right. Not 100% sure... If I were to guess, this would be my work
\[\large F_c=ma_c=\frac{mv^2}{r}=m(9g)\]If we divide out the m, then our equation is\[\large a_c=\frac{v^2}{r}\]And then plug in the numbers and see how it compares to 9g (88.2 \(m/s^2\))

Answer 5

Actually, cross out the m(9g). I don't think it's part of the equation -- however the information is given for us to compare probably.

Answer 6

So I would use the Ac = V^2/r formula so it would be 484^2/3470= 67.50893372. Then would I need to square root the 67.51 to get 8.22?

Answer 7

Why would you need to square root it?

Answer 8

Because the v is squared?

Answer 9

You're not solving for v, dude X)

Answer 10

I know.

Answer 11

We are solving for the acceleration.

Answer 12

Then... what are you saying anymore?

You already said that
484^2/3470= a_c

That's it. Don't know where you're getting the square root.

Answer 13

Then we use the extra information to compare whether or not the pilot can handle the acceleration. Whatever your answer is that you calculated, you need to see if it's greater than, less than, or equal to the max acceleration that the pilot can handle, which is given to you in the problem.

Answer 14

I thought we might have to but with acceleration being squared we wouldn't have to right?

Answer 15

Okay I get 67.51 m/s^2

Answer 16

\[\huge a_c=\frac{v^2}{r}\]Where is acceleration being squared?

Answer 17

I was thinking wrong. I thought we were needing to square root v^2 in order to get ac

Answer 18

I think you're making connections that aren't there. You probably just need to slow down and analyze what each piece of the equation is saying. X)

Answer 19

Yeah. We have V^2 and R.

Answer 20

V^2 = 485 m/s R=3470m

Answer 21

485^2/3470 =67.50893372m/s^2

Answer 22

Correction:
V^2 = 485 `^2` m/s R=3470m

But yes, so now compare that to 9g (9 times the acceleration due to gravity = 88.2 m/s^2)

Answer 23

the pilot wouldn't pass out. Okay the answer to the first part is 67.51 because I've been putting that and it's counting it wrong or do I need to do more?

Answer 24

I got 67.79. Not sure what the threshold is for the software that you use. I assumed it'd be close enough.

Answer 25

I got the same answer. I accidentally put 4 in for the 5 in 485. Thanks.

Answer 26

No problem X)