Question

A jet plane has a takeoff speed of vto = 75 m/s and can move along the runway at an average acceleration of 1.3 m/s2. If the length of the runway is 2.5 km, will the plane be able to use this runway safely?

Answer 1

You have two problems here, really. First, you need to see how long it will actually take the plane to GET to 75 m/s.

Try:
\[v = v_{0} + at\]

Here v0 is the initial velocity, a is acceleration, and t is time.

Answer 2

Solve for t.

Answer 3

Does it make sense? Here v0 is obviously 0, since the plane is starting from a stand still. V would be 75, because that's what we want our final velocity to be.

Answer 4

I solved for T = 58 seconds or (57.69)

Answer 5

Perfect. So it takes about 57.69 seconds for the plane to hit 75 m/s. So now the second part. See what distance 57.69 seconds amounts to with what we know:

\[x_{0} + v_{0}t + 0.5at^2\]

Here x0 is 0 (because we're starting from relative 0). V0 is still 0, since we're still starting from a stand still.

Answer 6

why isn't the length 2.5km or 2500 km?

Answer 7

That is *initial* starting length. We are starting from 0. We just want to see what length 57.69 seconds amounts to. We can compare it later.

Answer 8

So essentially x0 and v0t just go away.

Answer 9

excuse me 2500m

Answer 10

oh ok

Answer 11

So see what distance that amounts to using 1/2at^2

Answer 12

x = 2186 m

Answer 13

(0.5)(1.3)(57.69^2) = 2163m.
Yours is probably different due to round off, but close enough.

So the plane will consume 2186m of runway before hitting it's takeoff speed of 75 m/s. So this plane is good to go.

Answer 14

Since obviously 2186m < 2500m

Answer 15

Thank you very much, I was having difficulty seeing the problem. I am in you gratitude. Also I calculated 2163 as well using the orginal numbers.

Answer 16

No problem. Just try not to round off until the very end of the problem.