An aeroplane taking off from a field has a run of 500 meters. What is the acceleration and take off velocity if it leaves the ground 10 seconds after the start ?

Answer 1

vf=10*a
vf=sqrt(2ad)
or
10a=sqrt2ad
100a^2=2*500*a
a=10m/s^2

vf=10*10=100m/s

Answer 2

Assuming you know the relationship between distance, velocity and acceleration,

v = at , no constant since at t=0 , v=0
d = (1/2)at^2 , no constant since at t=0, d = 0

but when t = 10, d = 500
(1/2)a(100) = 500
a = 10

acceleration is 10 m/s^2,

Answer 3

Sam, it would have been nice if you had given bobpursley credit for his solution that you just cut-and-pasted.

http://www.jiskha.com/display.cgi?id=1402557272

Answer 4

Ahhh, yes. The age of copy/paste/plagiarize is well upon us!

That's a bad habit to get into, Sam. =(

Answer 5

is that a big deal? if it is , then im greatly sorry ....and could someone answer my question pls /display.cgi?id=1454771086