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 ?
vf=10*a
vf=sqrt(2ad)
or
10a=sqrt2ad
100a^2=2*500*a
a=10m/s^2
vf=10*10=100m/s
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,
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
Ahhh, yes. The age of copy/paste/plagiarize is well upon us!
That's a bad habit to get into, Sam. =(
is that a big deal? if it is , then im greatly sorry ....and could someone answer my question pls /display.cgi?id=1454771086
To find the acceleration and takeoff velocity of the airplane, we can use the equations of motion. The equations of motion relate displacement, time, initial velocity, final velocity, and acceleration.
First, let's define the variables:
- Initial velocity (u) = 0 m/s (since the airplane starts from rest)
- Displacement (s) = 500 meters
- Time (t) = 10 seconds
1. Calculate acceleration (a):
We can use the equation: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration.
Plugging in the values:
500 = 0 * 10 + (1/2) * a * (10^2)
Simplifying the equation:
500 = 0 + 5a * 100
500 = 500a
a = 1 m/s^2
Therefore, the acceleration of the airplane is 1 m/s^2.
2. Calculate takeoff velocity (v):
We can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values:
v = 0 + 1 * 10
Simplifying the equation:
v = 10 m/s
Therefore, the takeoff velocity of the airplane is 10 m/s.