Asked by shelly
A car traveling at 37 m/s breaks to a stop in 6.0 m and 3.5 s. What was the car’s acceleration?
Answers
Answered by
Elena
For decelerated motion
s= vt-at^2/2.
a=2(vt-s)/t^2=2(37•3.5-6)/3.5^2=20.16 m/s^2
s= vt-at^2/2.
a=2(vt-s)/t^2=2(37•3.5-6)/3.5^2=20.16 m/s^2
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