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
It is very hard to understand that how you have done this question please send step by step
To find the acceleration and takeoff velocity of the airplane, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s = displacement (run) of the airplane = 500 meters
u = initial velocity = 0 (since it starts from rest)
t = time = 10 seconds
Rearranging the equation, we have:
a = (2s - 2ut) / t^2
Substituting the given values:
a = (2 * 500 - 2 * 0 * 10) / (10^2)
a = 1000 / 100
a = 10 m/s^2
So, the acceleration of the airplane is 10 m/s^2.
Now, let's find the takeoff velocity. We can use the equation:
v = u + at
Substituting the values:
v = 0 + 10 * 10
v = 100 m/s
Therefore, the takeoff velocity of the airplane is 100 m/s.
To find the acceleration and takeoff velocity of the airplane, we'll use the kinematic equation:
v = u + at
Where:
v = final velocity (takeoff velocity)
u = initial velocity (0 m/s as it starts from rest on the ground)
a = acceleration
t = time (10 seconds)
First, let's find the acceleration:
To find the acceleration, we'll use the equation:
s = ut + (1/2)at^2
Where:
s = distance (500 meters)
u = initial velocity (0 m/s)
t = time (10 seconds)
a = acceleration (to be determined)
Rearranging the equation, we get:
a = (2s - 2ut) / t^2
Substituting the given values:
a = (2 * 500 - 2 * 0 * 10) / 10^2
a = 1000 / 100
a = 10 m/s^2
Therefore, the acceleration of the airplane is 10 m/s^2.
Now let's find the takeoff velocity:
Using the first equation:
v = u + at
Substituting the values:
v = 0 + 10 * 10
v = 100 m/s
Therefore, the takeoff velocity of the airplane is 100 m/s.