A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 1.80 km runway?
vf^2=2*a*distance solve for a
be certain to change velocity to m/s, and d to m
To find the least constant acceleration needed for takeoff, we can use the kinematic equation:
vf^2 = vi^2 + 2aΔx
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and Δx is the displacement. In this case, the initial velocity is 0 km/h (since the jet starts from rest), the final velocity is 360 km/h, and the displacement is 1.80 km.
First, we need to convert the velocities and displacement to meters per second (m/s) as it is the standard unit for acceleration.
1 km/h is equal to 0.2778 m/s. So, 360 km/h is equal to 100 m/s.
Similarly, 1.80 km is equal to 1800 m.
Now we can substitute the values into the equation:
(100 m/s)^2 = (0 m/s)^2 + 2a * 1800 m
Simplifying the equation:
10000 m^2/s^2 = 3600a m
Divide both sides by 3600 m:
2.78 m/s^2 = a
Therefore, the least constant acceleration needed for takeoff from a 1.80 km runway is 2.78 m/s^2.