A fully loaded 737 aircraft takes off at 250km/h.If its acceleration is a steady 3.3 , how long a runway is required? How much time does it take the plane to reach takeoff?
Can you clarify the units for the acceleration of 3.3?
Note: the acceleration of a 737 is in the range of 1.6-1.9 m/s².
3.3m/s^2
That is a little fast for a fully loaded 737, but I'm not buying one, so it doesn't matter!
Hint: convert all quantities to consistent units, in this case, in metres and seconds.
For example,
Final speed, v1
=250 km/hr = 250 *1000m / 3600 s
Acceleration, a
= 3.3 m/s²
Time required, t
= v1/a s
Length of runway required, S
= v1²/(2a)
Post your answers for a check if you wish.
where do you get the equation v1²/(2a)
To determine the length of the runway required for takeoff, we can use the kinematic equation:
distance = (initial velocity^2) / (2 * acceleration)
Given that the initial velocity is 250 km/h (which we need to convert to m/s), and the acceleration is 3.3 m/s², we can calculate the distance required for takeoff.
First, let's convert the initial velocity from km/h to m/s:
250 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 69.4 m/s (rounded to one decimal place)
Now, we can substitute the values into the kinematic equation:
distance = (69.4 m/s)^2 / (2 * 3.3 m/s²)
distance = 4800.0 m (rounded to the nearest meter)
Therefore, a runway length of approximately 4800 meters is required for takeoff.
To determine how long it takes the plane to reach takeoff, we can use another kinematic equation:
time = final velocity / acceleration
In this case, the final velocity is the takeoff speed (250 km/h) converted to m/s:
250 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 69.4 m/s (rounded to one decimal place)
Substituting the values into the equation:
time = 69.4 m/s / 3.3 m/s²
time = 21.0 s (rounded to one decimal place)
Therefore, it takes approximately 21 seconds for the plane to reach takeoff speed.