An object moving with a speed of 6.25 m/s, is deaccelerated at a rate given by ` a= - 2.5 sqrt(v) `, where v is the instantaneous speed. What is the time taken by the object to come to rest?

Explanation

Initial velocity of the lift u = 0 , acceleration a = 2`"m.s"^(-2)`
Let the rise of the lift in 4 s be s and its velocity at that point be v.
Hence from equation v = u+at we get,
v = 0 + ` 2xx4 = 8 "m.s"^(-1)`
Also from equation s = `ut + (1)/(2) at^(2)` we get,
`s = 0 xx 4 + (1)/(2) xx 2 xx (4)^(2) = 16 `m
`:.` The stone piece was dropped with an initial upward velocity of 8 `m.s"^(-1)` and was at a height of 16 m from the ground. If t is the time taken by the stone to reach the ground then from equation h = `ut + (1)/(2) "gt"^(2)` ,
`16 = - 8t + (1)/(2) xx 9.8t^(2)`
[`because` for the stone, u = - `8"m.s"^(-1)`, g = 9.8 `"m.s"^(-2)` , h = 16 m]
or, `" " 4.9t^(2) - 8t-16 =0`
or,`" " t=(8pmsqrt((8)^(2)-4xx4.9xx(-16)))/(2xx4.9)=(8 pm 19.4)/(9.8)`
Since time cannot be negative, `t = (8 +19.4)/(9.8)` 2.8 s.