Question

A jet airplane travelling at the speed of 500km/h ejects it's PRODUCTS of combustion at the speed of 1500km/h relative to the jet plane.what is the speed of the latter with respect to an observe in the ground

Answer 1

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Speed of the jet airplane, v jet = 500 km/h

Relative speed of its products of combustion with respect to the plane,

v smoke = – 1500 km/h

Speed of its products of combustion with respect to the ground = v′smoke

Relative speed of its products of combustion with respect to the airplane,

vsmoke = v′smoke – vjet

1500 = v′smoke – 500

v′smoke = – 1000 km/h

The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

I hope, this will help you

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Answer 2

Concept:—

★ Velocity of object A with respect to B.

Vab = Va-Vb

Velocity of object B with respect to A

Vba= Vb-Va

Velocity of object A with respect to ground

Va

Solution:—

speed of the jet airplane Vjet = 500 km/h

Relative speed of its products of combustion with respect to the plane,

→Vsmoke= V'smoke-Vjet

→ 1500= V'smoke-500

→ V'smoke= -1000 km/h

The negative sign indicates that the direction of its product of combustion is opposite to the direction of smoke.