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Mathematics 18 Online
OpenStudy (anonymous):

Calculus help on graph?

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 \(t=4\)

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)

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

OpenStudy (anonymous):

unless you have any spare force

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