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Solution 6.8c

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For the second stage where the lift slows down we first must calculate the maximum speed which is reached after 10 seconds using v=u+at.

From part a) we have the acceleration is 0.075 ms−2, giving the maximum sped reached is 0+0.07510=0.75 ms−1


use s=21(u+v)t. This gives the distance travelled during this stage is

s=210+0.755=1.875 m 

The total distance travelled using part b) is

3.75+1.875=5.625 m

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