Consider an elevator in a hospital. The elevator accelerates upward with constant acceleration for a distance of 100 cm and then starts to slow down. Also, it is known that the force exerted on a passenger by the floor of the elevator does not exceed 1. 8 times the passenger's weight. Now, find the maximum speed of the elevator in SI units: Take `g = 10 ms^(-2)`

Answer

`04.00`

Explanation

Consider the forces on the person:
`SigmaF_y=ma_y`
`R=mg=ma`
`R_("max") =1.8 mg`
`implies a_("max") =0.80 g = 8 ms^(-2)`
`v_("max")^2 =u^2 +2a_("max")`
`:. V_("max")^2 =0^2+2xx8xx1`
`:. v_("max")=sqrt16 =4m//s`