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CBSE Class 11-science Answered

A object moving with speed of 6.25 m/s, is decelerated  at a rate given by a=-2.5√v, where v is the instantaneous speed. The time taken by the object to come to rest would be
Asked by binayakbiswas | 16 Jul, 2019, 09:50: PM
answered-by-expert Expert Answer
Acceleration a = dv/dt = -2.5√v
 
 begin mathsize 14px style fraction numerator d v over denominator square root of v end fraction space equals space minus 2.5 space d t space space space space
B y space space i n t e g r a t i o n comma space space integral subscript v subscript i end subscript superscript v subscript f end superscript fraction numerator begin display style d v end style over denominator begin display style square root of v end style end fraction space equals space minus 2.5 integral subscript 0 superscript T d t
open vertical bar 2 square root of v close vertical bar subscript v subscript i end subscript superscript v subscript f end superscript space equals space minus 2.5 space T
sin c e space v subscript f equals 0 space a n d space v subscript i equals 6.25 space m divided by s space comma space space w e space g e t comma space space space space 2 square root of 6.25 end root space equals space 2.5 space T space space o r space space T space equals space 2 space s end style
 
Where T is time taken by the object to come to rest
Answered by Thiyagarajan K | 16 Jul, 2019, 10:49: PM
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