The front and back ends of a train moving with uniform aceleration pass a stationary observer with velocities 'u' and 'v' respectively. The velocity of the middle point of the train while passing the observer would be :
(1) √(u² + v²)
(2)√(u² + v²)/2
(3)(u² + v²)/2
(4)(u+v)/2
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The answer is option (4) (u + v)/2
Answered by
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the middle point of the train while passing the observer would be :
(4)(u+v)/2
(4)(u+v)/2
subhadippal7929:
how ??
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