An airplane taxis onto the runway going at 10 m/s.if it can accelerate at 3.0 m/s square and its take-off speed is 90 m/s, what length of runway will it need?
1333.33333....m which means equal to 1.33333km my response
1.3333 km
1333.33....m which means equal to
1.333..km
To find the length of the runway needed for the airplane to take off, we need to calculate the distance it will travel during its acceleration phase and add it to the distance it will cover at take-off speed.
First, let's calculate the time it will take for the airplane to reach its take-off speed:
We can use the formula:
v = u + at
where:
v = final velocity (take-off speed) = 90 m/s
u = initial velocity (taxi speed) = 10 m/s
a = acceleration = 3.0 m/s^2
t = time taken
Rearranging the formula to solve for t, we get:
t = (v - u) / a
t = (90 - 10) / 3.0
t = 80 / 3.0
t ≈ 26.67 seconds (rounded to two decimal places)
Next, let's find the distance covered during the acceleration phase:
We can use the formula:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity (taxi speed) = 10 m/s
t = time taken = 26.67 seconds (approx.)
a = acceleration = 3.0 m/s^2
Substituting the values into the formula, we get:
s = 10 * 26.67 + (1/2) * 3.0 * (26.67)^2
s ≈ 266.7 + 2000
s ≈ 2266.7 meters (rounded to one decimal place)
Finally, we can add the distance covered during the acceleration phase to the distance covered at take-off speed to find the total length of the runway needed:
Total length = distance during acceleration + distance at take-off speed
Total length = 2266.7 + 0 (since the distance covered at take-off speed is 0, as it happens instantaneously)
Total length ≈ 2266.7 meters
Therefore, the airplane will need approximately 2266.7 meters of runway to take off.