An airplane lands and starts down the runway at a southwest velocity of 48 m/s. What constant acceleration allows it to come to a stop in 1.5 km?

vf^2=vi^2 + 2ad solve for a

16 To find the constant acceleration that allows the airplane to come to a stop, we can use one of the kinematic equations:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement

In this case, we know that the initial velocity (u) is 48 m/s (southwest) and the final velocity (v) is 0 m/s (since the airplane is coming to a stop). The displacement (s) is given as 1.5 km, but it needs to be converted to meters for consistency.

1. Convert the displacement from kilometers to meters:
1.5 km = 1.5 * 1000 m = 1500 m

Now, we can plug in the given values into the equation:

0^2 = (48 m/s)^2 + 2a * 1500 m

Simplifying the equation:

0 = 2304 m^2/s^2 + 3000a m

To solve for the acceleration (a), we need to isolate it. Rearranging the equation:

-3000a m = 2304 m^2/s^2

Dividing both sides by -3000 m:

a = 2304 m^2/s^2 / -3000 m

a = -0.768 m/s^2

Therefore, the constant acceleration required for the airplane to come to a stop in 1.5 km is approximately -0.768 m/s^2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.