Posted by ricajane on Thursday, November 24, 2011 at 6:08am.
Given:
Vi = 12 m/s
a = - 2.5 m/s2
t = 5s
Required: Vf (final velocity)
Solution:
a=Vf^2 - Vi^2 / t
-2.5 m/s = Vf^2 - (12m/s)^2 / 5s
-2.5 m/s = Vf^2 - 144m^2/s^2
-2.5m/s over -144 m^2/s^2 = Vf^2
0.13 m/s = Vf
You must mean toll gate.
After decelerating 5 s at 2.5 m/s^2, it will have reduced its velocity by 12.5 m/s, and will be going backwards, at 0.5 m/s.
A rather strange situation.
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