6%D weBWork: 106-Fal Sign In or Sign Up I PDTU -Phi Deta Th W 11-4.8: Problem 1

Question

6%D weBWork: 106-Fal Sign In or Sign Up I PDTU -Phi Deta Th W 11-4.8: Problem 1 Problem List Next 1 point) The functions () and g() are shown below t the motion of a particle whose position at time f is given by x-f).yg) sketch a graph of the resulting motion and use your graph to answer the tollowing questions al The slope of the graph at (0.25,-0.5) is (enter undef id the slope is not defined 0% At this port te p cle is mon to the right c) The slope of the graph at(1.75-0.5) is ard, neither up nor down , enter undef if the slope is not definecd) d At this point the particle is maving nelther left nor right and neither up nor down Noter: You can earn partial credit on this problem Preview My Anowers Submit Answers atthempts remaining 3 4 5 6 8 0 0

Explanation / Answer

dy/dt is :

slope = (y2 - y1)/ (x2 - x1)

= (0.5) / (-0.25)

= -2 --> ANS

dx/dt is :
1

So, dy/dx= dy/dt divided by dx/dt

dy/dx = -2/1

= -2

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b)
Notice dy/dt is negative --< moving down
dx/dt is posiitve ---> rightward

So, right , moving down

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c)
When t= 1.75,
clearly dx/dt frm first graph = -1
and dy/dt from second graph passes thru (1.5,1) and (2,0) which has slope = -2

So, dx/dt = -1
and dy/dt = -2
S, slope of graph dy/dx = -2 / -1 ----> 2 ---> ANS

Also notice that both dy/dt and dx/dt are negative

So, moving left and moving down

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