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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?

Hi,
Here we will be considering the direction of velocity of the jet to be positive direction.
Given:
Speed of the jet w.r.t ground = 500 km/hr (t
Speed of the combustion product w.r.t jet =- 1500 km/hr (negative sign because it is in the direction opposite to the direction of velocity of the jet)
Therefore, applying the concept of relative velocity we get

Negative sign indicates that the velocity of the gas is in the direction opposite to the direction of the velocity of the jet.
Hope this helps,
Good luck.
 

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View Full Answer

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

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

vsmoke = – 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.

  • 70

The answer is there on meritnation Q's & A's. Physics, Chapt. Motion in a straight line.

  • -5

Chaap diya.....meritnation ke study form se, but i want the solution in another way......:-(

  • -4

 but this is also a correct way

  • -6

 as well as understandable. THen where's the doubt

  • -6

Hey u r in fifth na, ok then explain me the question in brief and correct way......!!!! x-(

  • -7

 I know it's from meritnations exercise solution, I did mention that. As far as another way is concerned, I;m afraid there is only one formula for us for relative velocity.

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 I m the users brother n I read in class X

  • -6

Thank u sir :-)

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