a) 12.5 m
Explanation:
During the acceleration phase, the distance travelled by the elevator is given by:
![v^2 - u^2 = 2ad](https://tex.z-dn.net/?f=v%5E2%20-%20u%5E2%20%3D%202ad)
where
v = 5 m/s is the final speed of the elevator
u = 0 is the initial speed
a = 1.0 m/s^2 is the acceleration
d is the distance travelled
Using the equation, we can find d:
![d=\frac{v^2-u^2}{2a}=\frac{(5 m/s)^2-0}{2(1.0 m/s^2)}=12.5 m](https://tex.z-dn.net/?f=d%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%3D%5Cfrac%7B%285%20m%2Fs%29%5E2-0%7D%7B2%281.0%20m%2Fs%5E2%29%7D%3D12.5%20m)
b) 45 s
Explanation:
The distance covered during the acceleration phase is 12.5 m. The distance covered during the de-celeration phase, at the end of the trip, is exactly the same (because the values of u, v and a are the same). The total distance is 200 m, so the distance covered at a constant speed of 5 m/s is
![d'=200 m- 12.5 m -12.5 m=175 m](https://tex.z-dn.net/?f=d%27%3D200%20m-%2012.5%20m%20-12.5%20m%3D175%20m)
This distance is covered at a speed of v=5 m/s, so the time taken is
![t'=\frac{d'}{v}=\frac{175 m}{5 m/s}=35 s](https://tex.z-dn.net/?f=t%27%3D%5Cfrac%7Bd%27%7D%7Bv%7D%3D%5Cfrac%7B175%20m%7D%7B5%20m%2Fs%7D%3D35%20s)
The time the elevator takes to cover the 12.5 m in the acceleration phase is
![t''=\frac{v-u}{a}=\frac{5 m/s-0}{1 m/s^2}=5 s](https://tex.z-dn.net/?f=t%27%27%3D%5Cfrac%7Bv-u%7D%7Ba%7D%3D%5Cfrac%7B5%20m%2Fs-0%7D%7B1%20m%2Fs%5E2%7D%3D5%20s)
and the same for the deceleration phase; therefore, the total time for the trip is
![t=t'+t''+t''=35 s+5 s+5s =45 s](https://tex.z-dn.net/?f=t%3Dt%27%2Bt%27%27%2Bt%27%27%3D35%20s%2B5%20s%2B5s%20%3D45%20s)