A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car?
We have equation of the motion
v=u+at
0= 22.4+2.55xa
-2.55a = 22.4 22.4
a -22.4 ÷2.55
a=-8.784m/s(square)
v2=u2+2as
0=(22.4)2+2*s*(-8.784)
-501.76=-17.568s
s=28.56m
There Before car stops it will travel 28.56 m.
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