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An object moving with a speed of 6.25 m/s , is decelerated at a rate given by: (dv)/(dt)= -2.5sqrt(v) where v is the instantaneous speedThe time taken by the object to come to rest would be: |
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Answer» 2s `:. (dv)/(sqrt(v)=-2.5 dt` or , `sqrt(v)` =- 2.5 =- 2.5 t +c: c=intergration constant. At t=0, v=6.25 as GIVEN in the question, puttings above `:. 2sqrt(6.25) =-2.5 xx0+c` or c=5.0 `:. 2sqrt(v)` = -2.5 t +5.0 When the OBJECT will COME to rest v will be zero putting above, `:.t=(5.0)/(2.5)=2s` |
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