.55 = Vi(2) + (1/2) a 2^2
0.55 = 2 Vi + 2 a
total distance after 4 seconds = .55+.77
= 1.32 m
1.32 = 4 Vi + (1/2) a (4^2)
1.32 = 4 Vi + 8 a
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so solve these two equations for Vi and a:
1.32 = 4 Vi + 8 a
0.55 = 2 Vi + 2 a
A remote-control vehicle moves forward with constant acceleration. During the first two seconds, it
travels 55.0(cm), and during the following two seconds, another 77.0(cm)
(a) Calculate the vehicle’s initial velocity and its acceleration.
(b) What distance will it travel the next 4.0(s)?
2 answers
thank you so much Damon!