Question

Someone is travelling at a velocity of 2.8e8 m/s away from a stationary observer and throws an object foward with a velocity of 2.1e8 m/s (relative to the student). find the velocity of the object with respect to the stationary observer.

Answer 1

Given , Speed of the person (v₁)= 2.8 x 10⁸ m/s and

Speed of the object (v₂) thrown forward = 2.1 x 10⁸ m/s

If you think classically then, velocity of the object with respect to the stationary observer will be :

v = (2.8 +2.1 ) x 10⁸ = 4.9 x 10⁸ m/s

Which is completely a wrong answer. Because it will violate the second postulate of Special Theory of Relativity . According to this postulate no object can move with a speed greater than the speed of light (C = 3 x 10⁸ m/s).

This type of velocity addition is done by using the most famous relativistic velocity addition formula . This formula is given by

   (this is derived from the special theory of relativity)

V) C2

So, velocity of the object with respect to the stationary observer will be

  

2.8 × 108+2.1 x 108 V) 2.8x 10° × 2·1x 108 (3x10

or,  

or,

4.9× 108 1 0.653333 V)

or,  

Hence, velocity of the object with respect to the stationary observer will be 2.96 x 10⁸ m/s .

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