An airplane must reach a speed of 185 mi/h to take off. If the runway is 467 m long, what is the minimum value of the acceleration that will allow the airplane to take off successfully? m/s2
I have been trying the formula vf^2=Vi^2+2AD and solving for A
SO my work has been
185^2=0^2+2A(467)
34225/934=A
36.643=A
I'm assuming my mistake is using speed as Vf but not sure how else to do it. TIA
Be careful of units.
You expressed the runway length in m (meters), but the takeoff speed is expressed in mi/h (miles/hour). You will need to do some units conversion.
You're on the right track with using the kinematic equation to solve for acceleration. However, to properly use this equation, you need to convert your measurements to consistent units.
First, let's convert the speed from miles per hour (mi/h) to meters per second (m/s). To do this, we can use the conversion factor of 1 mile = 1609 meters and 1 hour = 3600 seconds:
185 mi/h * (1609 m / 1 mi) * (1 h / 3600 s) ≈ 82.77 m/s
Next, let's convert the runway length from meters to kilometers to maintain consistency in units. There are 1000 meters in a kilometer, so we have:
467 m * (1 km / 1000 m) ≈ 0.467 km
Now, we can use the equation you mentioned:
vf² = vi² + 2ad
Where:
vf = final velocity (82.77 m/s)
vi = initial velocity (0 m/s, assuming the airplane starts from rest)
d = displacement (0.467 km = 467 m)
a = acceleration (to be solved)
Let's plug in the values we know and solve for the minimum acceleration required:
(82.77 m/s)² = (0 m/s)² + 2a * (467 m)
a = [(82.77 m/s)² - (0 m/s)²] / (2 * 467 m)
a = (6861.2429 m²/s²) / (934 m)
a ≈ 7.35 m/s²
The minimum acceleration required for the airplane to take off successfully is approximately 7.35 m/s².