Given a particle moving according to s(t)=t^2−10t+9?
Question
Find the total distance traveled during the first 8 seconds. (Hint: When is the particle traveling in the positive direction?
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General Physics
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Answer ( 1 )
s = t² – 10t + 9
v = ds/dt = 2t – 10
When t = 0:
s = 0² – 10*0 + 9 = 9m
v = 2*0 – 10 = -10m/s. (Particle is moving in negative direction.)
When t = 8s:
s = 8² – 10*8 + 9 = -7m
v = 2*8 – 10 = 6m/s. (Particle is moving in positive direction.)
So direction has reversed. This happens when v = 0, i.e. when 2t – 10 =0. So v = 0 when t = 5s.
When t = 5s:
s = 5² – 10*5 + 9 = -16m
From t = 0 to t = 5s the particle has moved from s = 9m to s = -16m (a distance of 25m). From t = 5s to t = 8s the particle has moved from s = -16m to s = -7m (a distance of 9m).
Total distance = 25 + 9 = 34m