t = 4.2
T = 8
a = 5
I guess it started from a dead stop.
x = (1/2) a t^2 during acceleration
Vi = speed at end of acceleration = a t
X = Vi T during constant speed so
distance = (1/2) a t^2 + Vi T
A car accelerates at a rate of 5m/s2 for a period of 4.2 seconds and then moves at a constant speed for 8 seconds more. How far did it travel?
2 answers
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