An airplane lands and starts down the runway at a southwest velocity of 43 m/s. What constant acceleration allows it to come to a stop in 1.1 km?
To solve this problem, we need to determine the constant acceleration required for the airplane to come to a stop. We can use the following kinematic equation to find the acceleration:
v^2 = u^2 + 2aS
Where:
v = final velocity (0 m/s, as the airplane comes to a stop)
u = initial velocity (43 m/s)
a = acceleration (unknown, what we're trying to find)
S = displacement (1.1 km = 1100 m)
By substituting the given values into the equation, we can solve for the acceleration:
0^2 = (43 m/s)^2 + 2a * 1100 m
Simplifying the equation, we have:
0 = 1849 m^2/s^2 + 2200a
Rearranging the terms, we get:
2200a = -1849 m^2/s^2
Now, we can solve for 'a':
a = (-1849 m^2/s^2) / 2200
Dividing the numerator by the denominator, we find:
a ≈ -0.8405 m/s^2
Therefore, the required constant acceleration for the airplane to come to a stop in 1.1 km is approximately -0.8405 m/s^2. Note that the negative sign indicates deceleration, since the airplane is slowing down.
Vf^2=Vi^2+2ad
Vf=0, given Vi, given d. solve for a.