Question

An object , moving with a speed of ` 6.25 m//s `, is decelerated at a rate given by :
`(dv)/(dt) = - 2.5 sqrt (v)` where `v` is the instantaneous speed . The time taken by the object , to come to rest , would be :

Answer 1

Given, (dv)/(dt) =- 2.5 sqrt v`
or ` (dv)/9sqrtv) =- 2.4 dt`
Let the object will come to resta after time (t), then velocity of object will be changing in time (t) from `0 625 ms^(-1)` rozero.
Integrating the above relation within the conditions of motion, we get
` int _(6.25) ^(0) (du)/(sqrtv) = int _(0)^(t) -2.5 dt or [2 sqrt 2]_(6.25)^(0) =- 2.5 [t] _(0)^(t)`
or ` 0- 2 sqrt 6.25 =- 2.5 t or -2 xx 2.5 =- 2.5 t`
or t= 2 s`.