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OpenStudy (anonymous):
OpenStudy (anonymous):
@satellite73
OpenStudy (anonymous):
that is the graph of the velocity?
OpenStudy (anonymous):
I think so.
OpenStudy (anonymous):
if so, since velocity is the same as speed if it is positive, the largest looks to be 3 at \(t=4\)
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OpenStudy (anonymous):
the net distance is the area under the curve, and the total distance is the area above the curve plus the area below the curve
OpenStudy (anonymous):
i guess that first part is supposed to be half a circle with radius 1, so that area is \(\frac{\pi}{2}\)
the area of the triangle you get via \(\frac{1}{2}bh\)
OpenStudy (anonymous):
since the base is 4 and the height is 3, the area under that large triangle is 6
the "area" under the small triangle is negative since it is below the x axis, and that is 1
OpenStudy (anonymous):
therefore net change is
\[\frac{\pi}{2}+6-1\] where as total distance is
\[\frac{\pi}{2}+6+1\]
OpenStudy (anonymous):
So the net force is 5 + pi/2 ? (thats the option given)
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OpenStudy (nincompoop):
net force?
OpenStudy (anonymous):
Opps, sorry I meant net distance lol. Im taking both physics and calculus at the same time so I typed that without thinking.
OpenStudy (anonymous):
net change, yes
OpenStudy (anonymous):
Thanks!
OpenStudy (anonymous):
may the change be with you
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