An airplane lands and starts down the runway at a southwest velocity of 41 m/s. What constant acceleration allows it to come to a stop in 1.3 km?
To find the constant acceleration required for the airplane to come to a stop, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the airplane comes to a stop)
u = initial velocity (41 m/s)
a = acceleration (unknown)
s = displacement (1.3 km = 1300 m)
We can rearrange the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the given values into the equation:
a = (0^2 - 41^2) / (2 * 1300)
a = (-1681) / 2600
a ≈ -0.646 m/s^2
Therefore, a constant acceleration of approximately -0.646 m/s^2 will allow the airplane to come to a stop in 1.3 km. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, causing deceleration.
Average velocity * time = stopping distance
(Vo/2) * t = X = 1300
Also, you can write Vo = a* t for the deceleration time.
The time t can be eliminated from the two equations, resulting in
(Vo/2)(Vo/a) = X
Vo^2 = 2 a X
Solve for a