The pilot stops the plane in 484 m using a constant acceleration of -8.0 m/s2. How fast was the plane moving before the braking began?
Vi = initial speed
484 = average speed during stop * t
484 = (Vi/2) * t
v = Vi + a t
0 = Vi -8 t
t = Vi/8
so
484 = (Vi/2)(Vi/8) = Vi^2 / 16
Vi^2 = 7744
Vi = 88 m/s
To determine the initial velocity of the plane before braking, we can use the following equation:
vf^2 = vi^2 + 2aΔx
where:
- vf is the final velocity (which is 0 m/s since the plane stops),
- vi is the initial velocity (what we need to find),
- a is the acceleration (-8.0 m/s^2),
- Δx is the distance covered during braking (484 m).
Plugging in the values into the equation, we have:
0^2 = vi^2 + 2(-8.0 m/s^2)(484 m)
Simplifying the equation further:
0 = vi^2 - 8.0 m/s^2 * 968 m
Rearranging the equation to solve for vi:
vi^2 = 8.0 m/s^2 * 968 m
vi^2 = 7744 m^2/s^2
Taking the square root of both sides of the equations, we have:
vi = √7744 m^2/s^2
vi = 88 m/s
Therefore, the plane was moving at a speed of 88 m/s before braking began.
To find the initial velocity of the plane before braking began, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the plane comes to a stop)
u = initial velocity (what we're trying to find)
a = acceleration (-8.0 m/s^2)
s = distance covered during braking (484 m)
Rearranging the equation, we have:
u^2 = v^2 - 2as
Substituting the given values, we get:
u^2 = 0^2 - 2(-8.0)(484)
Simplifying further:
u^2 = 0 + 7744
u^2 = 7744
Taking the square root of both sides:
u = √7744
Calculating this, we find:
u ≈ 88 m/s
Therefore, the plane was moving at approximately 88 m/s before the braking began.