An airplane can reach a takeoff speed of 45 m/s (from rest) in 7.0 seconds. What is the minimum runway length required for a safe takeoff?
Find acceleration:
acceleration= (final velocity-initial velocity)/time
Then plug into the distance formula:
The distance = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time)
Is the initial velocity 0?
yes, the key word from the problem: "from rest"
and is the final velocity 45/7?
no, it is the takeoff speed of 45 m/s
To find the minimum runway length required for a safe takeoff, we need to calculate the distance the airplane will travel during takeoff. This can be done using the equation:
Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2
In this case, the initial velocity is 0 m/s (since the airplane starts from rest), the time is 7.0 seconds, and we need to find the acceleration.
The acceleration can be calculated using the equation:
Acceleration = (Final velocity - Initial velocity) / Time
In this case, the final velocity is the takeoff speed of 45 m/s, the initial velocity is 0 m/s, and the time is 7.0 seconds.
Plugging these values into the equation, we get:
Acceleration = (45 m/s - 0 m/s) / 7.0 s = 6.43 m/s^2
Now we can substitute the values into the distance equation:
Distance = 0 m/s * 7.0 s + (1/2) * 6.43 m/s^2 * (7.0 s)^2
Calculating this equation, we find:
Distance ≈ 150.02 meters
So the minimum runway length required for a safe takeoff is approximately 150.02 meters.