A certain aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the braking force is a constant 445,000 N, what is the speed of the airplane 10.0 s later?

41.3 m/s

Explanation:

Speed: This can be defined as the rate of change of distance. The S.I unit of speed is m/s.

From Newton's fundamental equation,

F = ma Equation 1

Where F = force, m = mass, a = acceleration.

make a the subject of the equation,

a = F/m Equation 2

Given: F = 445000 N, m = 300000 kg.

Substitute into equation 2

a = 445000/300000

a = 1.48 m/s².

Also,

a = (v-u)/t Equation 3

Where v = final speed, u = initial speed, t = time.

Given: u = 27 m/s, t = 10 s.

Substitute into equation 3

1.48 = (v-27)/10

v-27 = 1.48×10

v-27 = 14.8

v = 41.3 m/s.

Thus the speed = 41.3 m/s

more than, less

less than, more

the right answer is 0.3 meters per second