An aircraft has a take-off speed 180km/h and has length 250m from the start just before take-off a serious snag is detected in the engine and the aircraft failed to take-off. There is a wall at a distance of 270 m from takeoff point
To calculate whether the aircraft will hit the wall or not, we need to determine how far the aircraft can travel from its initial position during the take-off attempt.
To do this, we can use the formula:
distance = speed x time
Given that the take-off speed is 180 km/h and the aircraft has a length of 250 m, we can convert the speed to meters per second (m/s) for consistency.
1 km/h is equivalent to 0.2778 m/s.
180 km/h x 0.2778 m/s = 50 m/s (approx.)
Let's assume it takes the aircraft t seconds to fail to take-off after it starts moving.
So, during that time, the aircraft will have traveled a distance of:
distance = speed x time
distance = 50 m/s x t
Now, we need to determine whether the distance covered by the aircraft is greater than the distance to the wall.
If the distance covered is greater than the distance to the wall, then the aircraft will hit the wall.
If the distance covered is less than or equal to the distance to the wall, then the aircraft will not hit the wall.
Given that the distance to the wall is 270 m, we can set up the following inequality:
50t ≤ 270
Divide both sides by 50:
t ≤ 5.4 seconds
Therefore, if the aircraft fails to take-off within 5.4 seconds, it will hit the wall.
Note: This calculation assumes that the aircraft maintains a constant speed during the attempted take-off. In reality, the situation might be more complex, but this calculation provides a rough estimate based on the given information.
To determine if the aircraft will hit the wall or not, we need to calculate its stopping distance based on its initial speed and deceleration.
First, let's convert the take-off speed from km/h to m/s:
180 km/h = (180 * 1000 m) / (60 * 60 s) = 50 m/s
Now, let's calculate the stopping distance. We can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the aircraft needs to stop)
u = initial velocity (50 m/s)
a = acceleration (negative value, as the aircraft is decelerating)
s = stopping distance (what we want to find)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the values into the equation:
s = (0^2 - 50^2) / (2a)
s = -2500 / (2a)
s = -1250 / a
Now, we need to find the acceleration. Since we're given the length of the aircraft from the start (250 m) and the distance to the wall (270 m), we can calculate the acceleration using the equation:
s = ut + (1/2)at^2
Where:
s = distance (270 m)
u = initial velocity (50 m/s)
a = acceleration (what we want to find)
t = time (what we want to find)
Substituting the values into the equation:
270 = 50t + (1/2)at^2
Simplifying the equation, we have a quadratic equation:
(1/2)at^2 + 50t - 270 = 0
We can solve this quadratic equation to find the time (t) and then substitute it back into the equation s = ut + (1/2)at^2 to find the acceleration (a).
Once we have the acceleration (a), we can substitute it back into the equation s = -1250 / a to find the stopping distance (s).
I hope this explanation helps you understand the steps involved in solving this problem.
Wrong question
d = 270m = Required stopping distance.
Vo = 180km/h = 180,000m/3600s = 50m/s. = Initial velocity before take-off.
V^2 = Vo^2 + 2a*d.
0 = (50)^2 + 2a*270
-540a = 2500
a = -4.63m/s^2.
V = Vo + a*t.
0 = 50 - 4.63t
t = ?.