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A skateboarder coasts down a long hill, starting from rest and moving with constant acceleration to cover a certain distance in 6 s. In a second trial, he starts from rest and moves with the same acceleration for only 2 s. How is his displacement different in this second trial compared with the first trial?

Answer :

lcmendozaf

Answer:

d1 = 9*d2  The second displacement is 9 times shorter.

Explanation:

We know that:  [tex]X = Vo*t + \frac{a*t^2}{2}[/tex]  where Vo = 0 for both trials

During the first trial:

[tex]d1 = \frac{a*t1^2}{2}[/tex] Since t1 = 6s:

[tex]d1 = 18a[/tex]

During the second trial:

[tex]d2 = \frac{a*t2^2}{2}[/tex] Since t2 = 2s:

[tex]d2 = 2a[/tex]

From these two equations we get:

d1 = 18/2*d2  =>  d1 = 9*d2

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