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 that allows the airplane to come to a stop, we can use the kinematic equation:
v² = u² + 2as
Where:
v is the final velocity (which is 0 in this case since the airplane comes to a stop),
u is the initial velocity (southwest velocity of 41 m/s),
a is the acceleration,
and s is the displacement (1.3 km = 1300 m).
Plugging these values into the equation, we have:
0² = (41 m/s)² + 2a(1300 m)
Simplifying further, we get:
0 = 1681 m²/s² + 2600a
Now, let's isolate the acceleration term:
2600a = -1681 m²/s²
Dividing both sides by 2600, we find:
a = -0.646 m/s²
Therefore, the constant acceleration required for the airplane to come to a stop in 1.3 km is approximately -0.646 m/s². The negative sign indicates that the acceleration acts in the opposite direction of the initial southwest velocity.
To find the constant acceleration needed for the airplane to come to a stop, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (0 m/s, since it comes to a stop)
u = initial velocity (41 m/s)
a = acceleration
s = displacement (1.3 km converted to meters)
Let's calculate the displacement in meters first:
1.3 km = 1.3 * 1000 = 1300 meters
Now, we can plug in the values into the equation and solve for acceleration (a):
0² = (41 m/s)² + 2a(1300 m)
Simplifying the equation:
0 = 1681 m²/s² + 2600a
Rearranging to solve for a:
2600a = -1681
a = -1681 / 2600
Calculating the value of a:
a ≈ -0.646 m/s²
Therefore, the constant acceleration required for the airplane to come to a stop in 1.3 km is approximately -0.646 m/s² (negative sign indicates deceleration).