Question

An object moving with a speed of `6.25 m//s`, is deceleration at a rate given by :
`(dv)/(dt) = -25 sqrt(v)` ,
where `v` is instantaneous speed. The time taken by the obeject, to come to rest, would be :
A. 1 s
B. 2 s
C. 4 s
D. 8 s

Answer 1

Correct Answer - B
We are given `(dv)/(dt) = -0.25 sqrt(v)` or `(1)/(sqrt(v)) dv = 2.5 dt`
On integrating, within limit `("at" t =0 , v_1 = 6.25 m s^-1)`
any at any time, `v_2 = 0`)
`:. underset(6.25 ms^-1)overset(0)(int) v^(1//2) dv = -2.5 underset(0)overset(t)(int) dt`
`2 xx[v^(1//2)]_(6.25)^0 = -(2.5) t rArr t = (-2 xx (6.25)^(1//2))/(-2.5) = 2s`.