The distance d=6m traveled under the constant acceleration a in time t=2.2s is given with the formula:
d=v0t−2at2 where v0=6m/s is the initial speed. Expressing the acceleration, obtain:
a=t2v0−t22da=2.22⋅6−2.222⋅6≈2.98m/s2 Answer. 2.98 m/s^2.
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