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Question 8

Distance =vt+(at^2/2) where v = initial velocity, t=time and a= accelation

v=0, t=4.9 and a= 5.47

Therefore, distance =0+(5.47*4.9^2)/2 = 65.667

Question 9

The athlete swims 50 m to one end of the pool and 50 m back hence travels a distance of 100 m. However his/her displacement, which is the net distance traveled is zero(0) meters since the athlete gets back to his/her origin which is a zero displacement.

Please let me know if you need any clarification. I'm always happy to answer your questions.