Question

A plane is traveling at 80 m/s. To prepare for landing, it decreases its speed to 50 m/s in 120 s.

Answer 1

Answer:

The deceleration of the plane is 0.25 m/$s^{2}$.

Explanation:

Deceleration is the converse of acceleration.

From first law of motion, we have;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Thus,

v = 50 m/s

u = 80 m/s

t = 120 s

50 = 80 + 120a

50 - 80 = 120a

-30 = 120a

a = $\frac{-30}{120}$

  = -0.25

a = -0.25 m/$s^{2}$

The deceleration of the plane is 0.25 m/$s^{2}$.