In order for the wright brothers' 1903 flyer to reach launch speed, it had to be acceleration uniformly along a track that was 18.3m long. A system of pulleys and falling weights provided the acceleration. If the flyer was initially at rest and it took 2.74 s for the flyer to travel the length of the track, what was the magnitude of its acceleration
To find the magnitude of the acceleration, we can use the equation of motion:
v = u + at
Where:
v = final velocity (launch speed)
u = initial velocity (0 m/s, as the flyer is initially at rest)
a = acceleration
t = time taken (2.74 s)
Since the initial velocity is 0 m/s, the equation simplifies to:
v = at
We can rearrange the equation to solve for acceleration:
a = v / t
We need to find the launch speed (v) first. The launch speed is the final velocity achieved by the flyer when it reaches the end of the 18.3 m track.
To find the launch speed, we can use the equation:
v^2 = u^2 + 2as
Where:
v = launch speed (final velocity)
u = initial velocity (0 m/s)
a = acceleration
s = distance traveled (18.3 m)
Since the initial velocity is 0 m/s, the equation simplifies to:
v^2 = 2as
We can rearrange the equation to solve for v:
v = √(2as)
Substituting the given values:
v = √(2 * a * 18.3)
Now we can substitute this value of v into the previous equation to find the acceleration.
a = v / t
a = (√(2 * a * 18.3)) / 2.74
Squaring both sides of the equation:
a^2 = (2 * a * 18.3) / (2.74)^2
a^2 = (36.6 * a * 18.3) / 7.5076
Multiply both sides by 7.5076:
7.5076 * a^2 = 36.6 * a * 18.3
Divide both sides by 36.6:
7.5076 * a^2 / 36.6 = a
Multiply both sides by 36.6:
7.5076 * a^2 = 36.6 * a
Rearranging the equation:
7.5076 * a^2 - 36.6 * a = 0
We can now solve this quadratic equation to find the magnitude of the acceleration.