A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.5 x 10 exponent 5 kg, its speed is 27.0 m/s and the net braking force is 4.30 x 10 exponent 5 N, (A) what id its speed 7.50 s later? (b) how far has it traveled in this time?

Question

A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass  is 3.5 x 10 exponent 5 kg, its speed is 27.0 m/s and the net braking force is 4.30 x 10 exponent 5 N, (A) what id its speed 7.50 s later? (b) how far has it traveled in this time?

drwls · Accepted Answer

First calculate the deceleration rate:

a = F/m = (4.30*10^5 N)/(3.5*10^5 kg)
= 1.229 m/s^2

(a) At that deceleration rate, the speed 7.50 s later is reduced by 7.5*1.229 = 9.2 m/s, making the speed 17.8 m/s at that time.

(b) For the distance traveled, multiply the AVERAGE speed by 7.50 s.
22.4 m/s * 7.5 s = ___ m