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

An object moving with a speed of 6.25 m/s, is deaccelerated at a rate

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