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 1.2 km?Answer is in ___ m/s2.

Vf^2=Vi^2+2a*distance

V = sqrt (2 a X)

X = 1200 m
V = 43 m/s

Solve for a. Add a minus sign because there must be deceleration. Or make the direction of the acceleration 180 degrees opposite from the landing direction.

To find the constant acceleration required for the plane to come to a stop, we can use the following equation of motion:

\(v^2 = u^2 + 2as\)

where:
- \(v\) is the final velocity (0 m/s as the plane comes to a stop)
- \(u\) is the initial velocity (43 m/s)
- \(a\) is the constant acceleration we are trying to find
- \(s\) is the displacement (1.2 km = 1200 m)

By substituting the given values into the equation, we can solve for \(a\):

\(0^2 = 43^2 + 2a(1200)\)

Simplifying the equation:

\(0 = 1849 + 2400a\)

Rearranging the equation:

\(2400a = -1849\)

Finally, solving for \(a\):

\(a = \frac{{-1849}}{{2400}}\)

Therefore, the constant acceleration required for the plane to come to a stop in 1.2 km is approximately -0.7704 m/s².

To find the constant acceleration required for the airplane to come to a stop, we first need to convert the distance from kilometers to meters.

Given:
Initial velocity (u) = 43 m/s
Distance (s) = 1.2 km = 1200 m

We can now use the equation of motion that relates acceleration (a), initial velocity (u), distance (s), and final velocity (v) to find the required acceleration:

v^2 = u^2 + 2as

Since the airplane needs to come to a stop, the final velocity (v) is zero. Therefore, the equation becomes:

0 = (43)^2 + 2a(1200)

Rearranging the equation, we get:

0 = 1849 + 2400a

Simplifying, we have:

2400a = -1849

Dividing both sides by 2400, we find:

a = -1849 / 2400

Therefore, the required constant acceleration to bring the airplane to a stop in a distance of 1.2 km is approximately -0.7704 m/s^2.