An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects of air resistance, friction from the tires, and reverse thrust from the engines produce a constant force of 2.90×105N2.90×105N opposite to the airliner's motion. What distance along the runway does the airliner travel before coming to a halt?

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

Resistive forces = (2.90 × 10⁵) N = 290000 N

Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

Velocity of airliner = 75 m/s

Let the distance over moved by the airliner be equal to d

According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

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