Question

A plane landing on a small tropical island has
just 64 m of runway on which to stop.
If its initial speed is 66 m/s, what is the
maximum acceleration of the plane during
landing, assuming it to be constant?
Answer in units of m/s
2


How long does it take for the plane to stop
with this acceleration?
Answer in units of s

Answer 1

help please

Answer 2

@Agent47 help

Answer 3

This looks familiar. Didn't you already do this?

Answer 4

i got it wrong

Answer 5

well the second part

Answer 6

OK, I'll take a closer look at it and see if I can show you the setup in a clear way.
Give me a few seconds.

Answer 7

thanks

Answer 8

The acceleration you got last time, -34.03125 m/s^2 is correct, and I'm getting 64/33 s for the time to stop using two different formulas.

Answer 9

yea i got -34.03125 but whats the 64/33

Answer 10

64/33 ≈ 1.94s

Answer 11

Though that should be rounded to 1.9s given the limits of precision you have.

Answer 12

that answer was wrong

Answer 13

hmmm..
The acceleration is found by:
\[(v_f)^2-(v_o)^2=2a(x_f-x_o)\]
\[(0m/s)^2-(66m/s)^2=2a(64m-0m) \rightarrow\]
\[a=\frac{(0m/s)^2-(66m/s)^2}{2(64m-0m)} \space \space = -34.03125m/s^2\]

Answer 14

The time can be found by either
\[x_f=x_o+v_ot+\frac{1}{2}at^2 \space \space or \space \space v_f-v_o=at\]

Answer 15

the time part confuses me

Answer 16

\[64m=0m+66m/s(t)+\frac{1}{2}(-34.03125m/s^2)(t)^2 \space \space or\]
\[0m/s−66m/s=(-34.03125m/s^2)(t)\]

Answer 17

Using either of those equations,
\[t=1\frac{31}{33}≈1.939393...\]

Answer 18

divide -66 by -34.01325 and that equals t. time

Answer 19

Yes.

Answer 20

1.939393

Answer 21

thanks i got the answer by myself the first time but i dont know what happend thank you very much

Answer 22

It all checks out. I don't know why you're getting errors. Personally, with all the uncertainty, I would round that up to 2 seconds.

Answer 23

yea thanks i know now