Question

Question 3.5 A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on ground?

Class XI Physics Motion In A Straight Line Page 56

Answer 1

hello !
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Here , 
Speed of jet airplane (VJ) = 500 km /h 
Relative speed of gases (VGJ) = 1500 km/h 
Let Speed Of Gases = (VG) 
Also , 
VGJ = VG + VJ 
1500 = VG + 500 
VG   = 1000 km/ h 

Let the Speed of observer = VO 
But Speed of observer = 0 km/hr because he is at rest 

SO , VGO = VG - VO 
                 = 1000 - 0 
                 = 1000 km/hr 

Answer 2

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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,

v(smoke) = v′(smoke) – v(jet)

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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