Question

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 1

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

Step-by-step explanation:

We know that:
`X = Vo*t + (a*t^2)/(2)` where Vo = 0 for both trials

During the first trial:

`d1 = (a*t1^2)/(2)` Since t1 = 6s:

`d1 = 18a`

During the second trial:

`d2 = (a*t2^2)/(2)` Since t2 = 2s:

`d2 = 2a`

From these two equations we get:

d1 = 18/2*d2 => d1 = 9*d2