A Cessna 150 aircraft has a lift-off speed of
approximately 125 km/h.
What minimum constant acceleration does
this require if the aircraft is to be airborne
after a take-off run of 172 m?
Answer in units of m/s2
What is the corresponding take-off time?
Answer in units of s
If the aircraft continues to accelerate at this
rate, what speed will it reach 38.5 s after it
begins to roll?
Answer in units of m/s
To find the minimum constant acceleration required for the aircraft to be airborne after a take-off run of 172 m, we can use the following kinematic equation:
\(v_f^2 = v_i^2 + 2a d\)
where:
- \(v_f\) is the final velocity (lift-off speed) = 125 km/h = 34.7 m/s
- \(v_i\) is the initial velocity (assumed to be 0 m/s since it starts from rest)
- \(a\) is the acceleration (to be found)
- \(d\) is the distance traveled = 172 m
Substituting the given values into the equation, we have:
\(34.7^2 = 0^2 + 2a(172)\)
Simplifying the equation, we get:
\(1204.09 = 344a\)
Now, we solve for \(a\) by dividing both sides of the equation by 344:
\(a = \frac{1204.09}{344} \approx 3.50\) m/s²
Therefore, the minimum constant acceleration required for the aircraft to be airborne is approximately 3.50 m/s².
To find the corresponding take-off time, we can use the equation of motion:
\(v = v_0 + at\)
where:
- \(v\) is the final velocity (lift-off speed) = 34.7 m/s
- \(v_0\) is the initial velocity (0 m/s)
- \(a\) is the acceleration (3.50 m/s²)
- \(t\) is the time we need to find
Rearranging the equation, we have:
\(t = \frac{v - v_0}{a}\)
Substituting the given values, we get:
\(t = \frac{34.7 - 0}{3.50} \approx 9.91\) s
Therefore, the corresponding take-off time is approximately 9.91 seconds.
To find the speed of the aircraft 38.5 seconds after it begins to roll, we can use the formula:
\(v = v_0 + at\)
where:
- \(v\) is the final velocity (to be found)
- \(v_0\) is the initial velocity (0 m/s)
- \(a\) is the acceleration (3.50 m/s²)
- \(t\) is the time = 38.5 s
Substituting the given values, we have:
\(v = 0 + 3.50(38.5)\)
Calculating the equation, we get:
\(v \approx 134.75\) m/s
Therefore, the speed of the aircraft 38.5 seconds after it begins to roll is approximately 134.75 m/s.