Question

Help please! A plane traveling at 300.0 km/h attempts to land on a 500 m runway. The plane’s engines and brakes accelerate uniformly at –5.0 m/s2. Will the plane be able to land safely? Show all your work. Include any equations used, the given and unknown. Show all calculations. Express numerical answers with the correct number of significant digits. Remember to carry one or two extra decimal places throughout calculation. Do not round until the end

Answer 1

@DemolisionWolf

Answer 2

wait, are you in a physics class using kinematic equations? I can' rmember

Answer 3

yes

Answer 4

ok ok ok ^_^
so the first thing we need to do is identify which of the 3 kinematic equations we need to use.
which one to use depends on what we know.
we know:
initial velocity
change in distance
and -acceleration
look over the 3 kinematic equations, and pick which one you think is best

Answer 5

um d=vi*t+1/2*a*t^2?

Answer 6

awh, that was my first choice too, but then i was looking at it and it has time, and i'm not sure how i'd go about the problem using time. so, scratch that 1 out. so, now you have 2 left to pick from

Answer 7

vf^2=vi^2+2ad?

Answer 8

thats what I picked ^_^
so plug in all that we know, and see what vf is

Answer 9

remember to convert your speed to m/s from the km/h given

Answer 10

vf^2=83.3^2 + 2(-5)(500)?

Answer 11

correct!

Answer 12

vf^2 = 6938 - 5000
vf^2 = 1938
vf = sqrt(1938)
vf = 44.0227 m/s
so at the end of the 500m run way the plane has gone from an initial velocity of 83.33m/s to 44.02m/s
do you 'see it'?

Answer 13

yes

Answer 14

@washcaps
there is another way to solve it, or to look at it.
the first way we just did, looked at finding the final velocity of the plane after braking for 500m.
the second way you can figure it out is to solve for how long the run way would need to be in order to stop on it. this way would look like:
0^2 = 83.3^2 + 2*(-5)*d
then solve for d
which d = 695m
and 695 is much larger than the 500m runway we have, so the plane wont stop in time.

so this is another way of looking at this problem
^_^

Answer 15

Wow ok thanks man! Your great help and really explain the problems well

Answer 16

your welcome