An airplane lands and starts down the runway at a southwest velocity of 55 m/s. What is the magnitude of constant acceleration that allows it to come to a stop in 1.00 km

-1.5m/s²

Consider one of the equations of motion as follows;

v² = u² + 2as     -------------------------(i)

Where;

v = final velocity of moving body (airplane in this case)

u = initial velocity of the body

a = acceleration of motion of the body

s = distance covered by the body

From the question;

v = 0             [since the airplane comes to a stop]

u = 55m/s

s = 1.00km = 1000m

Substitute these values into equation (i) as follows;

0² = 55² + 2a(1000)

0 = 3025 + 2000a

Collect like terms;

2000a = -3025

Solve for a;

a =

a = -1.5m/s²

Therefore the acceleration that allows the plane to come to a stop in 1.00km is -1.5m/s².

The negative sign shows that the plane is actually decelerating.

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5