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Andrews [41]
4 years ago
7

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?
Physics
1 answer:
VashaNatasha [74]4 years ago
5 0

Answer:

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

Explanation:

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

During the first trial:

d1 = \frac{a*t1^2}{2} Since t1 = 6s:

d1 = 18a

During the second trial:

d2 = \frac{a*t2^2}{2} Since t2 = 2s:

d2 = 2a

From these two equations we get:

d1 = 18/2*d2  =>  d1 = 9*d2

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What force is required to accelerate to 10 kg object to 5.9 m/s/s?
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