Question 1066021: in a time of t hours, a particle moves a distance of s kilometers from its starting point, where s=e^t*tan(t)
find the average velocity between t = o.25 and t = 0.25 + h if
h = 0.01
h = 0.001
h = 0.0001
can somebody help me with this by showing me how to do it and explain it?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! average velocity = Net displacement / Total time
:
I assume the particle is moving in a straight line, otherwise, one has to deal with projections from the particle to the initial line defined at time 0.25
:
for t = 0.25, the particle is at position
s = e^(0.25*tan(0.25)) = 1.001091433 Km
:
for h = 0.01, t = 0.25 + 0.01 = 0.26
s = e^(0.26*tan(0.26)) = 1.001180547 Km
average velocity = (1.001180547 - 1.001091433) / (0.01) = 0.0089114 Km / hour
:
for h = 0.001, t = 0.25 + 0.001 = 0.251
s = e^(0.251*tan(0.251)) = 1.001100187 Km
average velocity = (1.001100187 - 1.001091433) / (0.001) = 0.008754 Km / hour
:
for h = 0.0001, t = 0.25 + 0.0001 = 0.2501
s = e^(0.2501*tan(0.2501)) = 1.001092307 Km
average velocity = (1.001092307 - 1.001091433) / (0.0001) = 0.000874 Km / hour
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