Question

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v = 288 m/s. Upon landing, the plane can produce an average deceleration of a = 19 m/s2.

1. How long will it take the plane, in seconds, to circle the Earth at the equator?

2. What is the numeric value for the minimum landing distance, d (in meters), this plane needs to come to rest?

Answer 1

We all know the relationship between velocity-distance-time
\[v=\frac{ x }{ t }\]

Answer 2

So if we substitute with
distance x=6,370 km=6,370,000 m
and
velocity v=288 m/s
we can get the time to circle the earth at the equator
\[t=\frac{ x }{ v }=\frac{ 6370000 }{ 288 }=22118\] seconds

Answer 3

For the second part:

Answer 4

wait, it says it's wrong. I think we have to square the distance since 6370 is radius, right?

Answer 5

We know the law of motion that states :
\[v_{f}^{2}=v_{0}^{2}-2*a*d\]
where a is the deceleration and d is the minimum distance to stop completely

Answer 6

so
\[288^{2}=0^2-2*19*d\]
therefore \[d=2182\] m

Answer 7

wait, it says the answer for the first part is wrong

Answer 8

I see
Thank you.
I will correct it now.
Good to see others like you revise what you write

Answer 9

For the first part, we would calculate the perimeter of the earth that the plane will circle it.
Earth perimeter/circumference = \[\pi(r)^2=\pi(6,370,000)^2=1.275274\times 10^{14}\]
meter

Answer 10

then the time required to circle the earth at the equator is
\[t=\frac{ x }{ v }=\frac{ 1.275274\times 10^{14}/288}\]
=\[4.428\times 10^{11}\] seconds

Answer 11

Hope that helps you.