An elevator is stopped at the ground floor. It starts moving upwards at constant acceleration a>0 for 5 seconds. It then keeps a constant speed for 35 seconds. Finally, it slows down with an acceleration of the same magnitude (but opposite direction) −a , until it comes to a halt at the top floor. The top floor is 320 meters above the ground floor.

Question

(a) What is the maximal speed v of the elevator ? (in m/s)

(b) What is the acceleration a ? (in m/s2)

Answer 1

The average speed during the acceleration, and deacceleration is V/2. Theconstant velocity is V.

distance=V(t-10)+V/2 * 10

solve for V.

the distance during accelerating is V/2*5

Vf^2=vo^2+2ad

where d=avg velocity*time=V/2 * 5
Vf^2=Vo^2+2ad
V^2/(v*5)=a=V/5
check that.

Answer 2

Hello

Can you explain the first equation?

distance=V(t-10)+V/2 * 10

Where is this distance coming from for me to be able to solve for V?

Also where is the 10 coming from in V(t-10) and V/2*10?

My final equation solving for V is :

V= 10d / 4(t-10)