A pilot stops a plane in 484 m using an acceleration of -8 m/s2. How fast was the plane moving before the breaking began? Thanks.
V = sqrt(2 a X)
where a is the deceleration rate and X is the distance
89 km/h at 54 north of east
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 stops)
u = initial velocity (what we're trying to find)
a = acceleration (-8 m/s^2)
s = distance covered (484 m)
Plugging in the given values into the equation, we have:
0^2 = u^2 + 2(-8)(484)
u^2 = -2(-8)(484)
u^2 = 7744
Taking the square root of both sides, we get:
u = √7744
u ≈ 88 m/s
Therefore, the initial velocity of the plane before braking began was approximately 88 m/s.
To find the velocity of the plane before the braking began, you can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, as the plane stops)
u = initial velocity (what we need to find)
a = acceleration (-8 m/s^2)
s = displacement (484 m)
Rearranging the equation, we have:
u^2 = v^2 - 2as
Substituting the known values into the equation:
u^2 = 0^2 - 2(-8)(484)
Simplifying the equation further:
u^2 = 7744
Taking the square root of both sides:
u = √7744
u ≈ 88 m/s
Therefore, the plane was moving at approximately 88 m/s before the braking began.