A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 14 m above the ground. If the elevator can have maximum acceleration of `0.2 m//s^2` and maximum deceleration of `0.1 m//s^2` and can reach a maximum speed of 2.5 m//s, determine the shortest time to make the lift, starting from rest and ending at rest.

Answer

B

Explanation

Area of v-t graph =displacement

`:. 1/2(3t)(0.2t)=14`
Solving this equation, we get
`3t=20.5s`