An object moving with the speed of 6.25m/s, is decelerated at a rate given by:dv–2.5vdt?where v is the instantaneous speed. The time taken by the object, to come to rest would be?

Question

Vikas TU · Answer

dv = – 2.5 v dt integerating both sides, dv/v = -2.5dt logv = -2.5t to come to rest velcity should be zero and log0 = – infinity therefore, from the concept the object would never stops .

dhiraj Raj · Answer

Dv/√v=-2.5dt integrating both side, 2√v=-2.5t+k at t=0, v=6.25 2*2.5=k=5 therefore for v=0, t=2 sec............

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An object moving with the speed of 6.25m/s, is decelerated at a rate given by:dv–2.5vdt?where v is the instantaneous speed. The time taken by the object, to come to rest would be?

An object moving with the speed of 6.25m/s, is decelerated at a rate given by:dv–2.5vdt?where v is the instantaneous speed. The time taken by the object, to come to rest would be?

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An object moving with the speed of 6.25m/s, is decelerated at a rate given by:dv–2.5vdt?where v is the instantaneous speed. The time taken by the object, to come to rest would be?

An object moving with the speed of 6.25m/s, is decelerated at a rate given by:dv–2.5vdt?where v is the instantaneous speed. The time taken by the object, to come to rest would be?

2 Answers

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