Question
A motorcycle starts from rest witha constant acceleration of 0.10m/s^2. After 30 s, the bike has its acceleration doubled( assume instantaneously ). How fast is it moving after a total distance of 200 m
Answers
Steve
initial distance is (1/2)(.10)(30^2) = 45
at that point, v = at = 3 m/s
so, after the bump in acceleration,
s = 45 + 3t + (1/2)(.2)t^2
when s=200, we have
45+3t+.1t^2 = 200
t = 27.131
by then, v = 3 + .2*27.131 = 8.43 m/s
at that point, v = at = 3 m/s
so, after the bump in acceleration,
s = 45 + 3t + (1/2)(.2)t^2
when s=200, we have
45+3t+.1t^2 = 200
t = 27.131
by then, v = 3 + .2*27.131 = 8.43 m/s