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 speed The time taken by the object to come to rest would be:

Options

A. 2s
B. 4 s
C. 8 s
D. 1 s

Answer

a

Explanation

Here `(dv)/(dt)=-25sqrt(v)`
`:. (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`