An engineer must design a runway to accommodate airplanes that must reach a ground velocity of 60 m/s before they can take off. These planes are capable of being accelerated uniformly at the rate of 2.1 m/s2.
(a) How long will it take the planes to reach takeoff speed?
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(b) What must be the minimum length of the runway?
To find the answers to these questions, we will use the equations of motion, specifically the equation for constant acceleration:
v = u + at
where:
- v is the final velocity (takeoff speed) in m/s,
- u is the initial velocity (ground velocity) in m/s,
- a is the acceleration in m/s^2,
- t is the time taken to reach the final velocity in seconds.
Let's solve these step by step:
(a) How long will it take the planes to reach takeoff speed?
We are given:
- Initial velocity (u) = 0 m/s (since the planes start from rest on the ground)
- Final velocity (v) = 60 m/s
- Acceleration (a) = 2.1 m/s^2
Using the equation v = u + at, we can rearrange it to solve for time (t):
t = (v - u) / a
Substituting the given values, we have:
t = (60 - 0) / 2.1
t = 60 / 2.1
t ā 28.57 seconds
Therefore, it will take approximately 28.57 seconds for the planes to reach takeoff speed.
(b) What must be the minimum length of the runway?
To find the minimum length of the runway, we need to use the equation for distance covered:
s = ut + (1/2)at^2
where:
- s is the distance covered (runway length)
- u is the initial velocity (ground velocity) = 0 m/s
- t is the time taken to reach takeoff speed ā 28.57 seconds
- a is the acceleration = 2.1 m/s^2
Plugging in the values, we get:
s = (0)(28.57) + (1/2)(2.1)(28.57)^2
s = 0 + (1/2)(2.1)(816.91)
s ā 874.68 meters
Therefore, the minimum length of the runway should be approximately 874.68 meters. This is the distance needed for the planes to accelerate uniformly and reach takeoff speed of 60 m/s.