an aeroplane travels 280m down the runway before taking off.if it start from rest , moves constant acceleration and becomes airbone in 8 second what is its speedin m/s,when it takes off
vf=at or vf=a*8
and..
vf^2=vi^2 + 2ad
vf^2=2a*280
so one has two equations..
vf=8a and
vf^2=560a
vf=8(vf^2/560) or
vf=560/16 check my math...
To solve this problem, we can use the equation of motion known as the SUVAT equation. The SUVAT equation relates the initial velocity (u), final velocity (v), acceleration (a), time taken (t), and displacement (s). In this case, we need to find the final velocity when the airplane takes off.
Given:
Displacement (s) = 280 m
Acceleration (a) = ?
Time (t) = 8 s
Initial velocity (u) = 0 m/s (since it starts from rest)
Final velocity (v) = ?
The SUVAT equation we will use is:
s = ut + 1/2at²
Since the airplane starts from rest, the initial velocity (u) is zero. The equation becomes:
s = 0 + (1/2)a(8²)
280 = 4a(64)
280 = 256a
To find the acceleration (a), we can rearrange the equation:
a = 280 / 256
a ≈ 1.09375 m/s²
Now, we can use the final velocity equation (v = u + at) to find the final velocity when the airplane takes off:
v = 0 + (1.09375)(8)
v ≈ 8.75 m/s
Therefore, the speed of the airplane when it takes off is approximately 8.75 m/s.