Question

skater goes down mountain for 6s, starts again from same point on mtn. but only for 2s. (constant acceleration) . would the second velocity be 1/9?

Answer 1

First trial is V1=g(6s), second trial is V2=g(2s). if V1/V2= g(6s)/g(2s) so it should be 3 or 1/3 if the reciprocal rate is taken. There is no mention of any inclination in the mountain. Initial V in both is 0.

Answer 2

Thank you. It was a relationship problem like you assumed.

Answer 3

If X1=X(0)+Vi(0)+g(6s^2)/2 then X1= g36/2 and
X2=X(0)+Vi(0)+g(2s^2)/2 then X2= g 4/s then
X1/X2 would be (g36/2)/(g4/2) = 9 or if the reciprocal rate is taken 1/9 but now we are only taking of displacement rate, not velocity.

Answer 4

The question asks about displacement. I was trying to shorten the problem and I made it vague. It is 1/9 then