Question

a 747 jetliner lands and begins to slow to a stop as it moves along the runway. if its mass is 3.50 x 10^5 kg. it speed is 75.0 m/s and the net breaking force is 7.25 x 10^5 N, what is its speed 10.0 seconds later? how far has it traveled in this time?

Answer 1

• The speed after 10 s is 54.29 m /s.

• The distance traveled at that time is 646 m.

Step-by-step explanation:

First calculate the deceleration rate:

a = F/m

= (7.25
`*` 10^5 N) / (3.5
`*` 10^5 kg)

a = 2.071 m/s^2 .

• At that deceleration rate, the speed after 10 s later is reduced by,

speed = 10
`*` 2.071

= 20.71 m / s making the speed as 54.29 m/s at that time.

(i.e. subtracting the two speeds = 75 m/s - 20.71 m/s = 54.29 m / s.)

• For the distance traveled, multiply the average speed by 10 s.

Average speed =( 75 m/s + 54.29 m/s) / 2 = 64.645 m / s.

distance traveled = average speed
`*` time

= 64.645
`*` 10

distance traveled = 646 m.