Question

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

Answer 1

Answer:

a = -1.51 m/s^2

Explanation:

Given:

Vi= 55 m/s

Vf= 0 m/s

S = 1km =1000 m

a = ? m/s^2

Sol:

3rd equation of motion

2as = vf^2 - vi^2

a = (55)^2 / 2x1000

a = -1.51 m/s^2

-'ve sign shows the deceleration.

Answer 2

Answer:

-1.5m/s²

Explanation:

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 = $\frac{-3025}{2000}$

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.