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 :

An object , moving with a speed of ` 6.25 m//s `, is decelerated at a

1.	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» 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`.

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

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