A hotel elevator ascends 100 m with a maximum speed of 4.1 m/s. Its acceleration and deceleration both have a magnitude of 1.2 m/s^2


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Velocity with accelration is calculated from

V = AT

where V is velocity, A is acceleration, T is time. You want the velocity to be its maximum (4.1 m/s). You know the acceleration, so you can calculate the time to do both the acceleration and deceleration.

The distance traveled during acceleration and deceleration is then given by this equation:

D = \frac 12 AT^2

where D is the distance traveled, A is the acceleration, and T is the time. You accelerate from 0 to 1.2 m/s^2.

Calculate the distance traveled during acceleration and deceleration and subtract that from the total distance traveled. You know the speed for this middle section (4.1 m/s), so figure out the time.

D = VT

D = distance, V = velocity, T = time.

You have also calculated the time for the acceleration an deceleration, so add those to the time to get the total time.