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OpenStudy (anonymous):
OpenStudy (mrhoola):
s=490t^2 represents a funtion of time yields the objects height (position) -correct?
OpenStudy (anonymous):
yes
OpenStudy (mrhoola):
if they want to know how much time it take for the object to fall the first 160 cm . which variable do we use to to plug in this number
???
OpenStudy (anonymous):
plug it into t?
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OpenStudy (mrhoola):
t rperesents time
No
OpenStudy (anonymous):
make equal to. 160=490t^2
OpenStudy (mrhoola):
Very Good! yes. now solve for time !
OpenStudy (anonymous):
t=4/7. what about the second part of A?
OpenStudy (mrhoola):
what do they mean by period?
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OpenStudy (anonymous):
i think it's for the time period
OpenStudy (mrhoola):
at 4/7 seconds?
OpenStudy (anonymous):
yes
OpenStudy (mrhoola):
or zero seconds to 4/7 seconds
OpenStudy (anonymous):
the answer to that is 280 cm/ sec. but i don't know how to get there
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OpenStudy (mrhoola):
ohh okay. so the Original function that was given represents position, if we take the derivative of the function it will give us the velocity of the object at any time 't'
OpenStudy (anonymous):
s'(t)=980t
OpenStudy (mrhoola):
yes
that is correct
OpenStudy (mrhoola):
now plug in the time we got in part 'a'to determine the instaneious velocity
OpenStudy (anonymous):
560
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OpenStudy (mrhoola):
hmmm . that is not the answer huh?
OpenStudy (anonymous):
or plug it into s=490(4/7)
to get 280
OpenStudy (mrhoola):
thats incorrect be cause it should be (4/7) ^ 2 - right ?
OpenStudy (anonymous):
yeah
OpenStudy (mrhoola):
isnt average velocity an equation described as ; distance divided by the interval of time (aka period)
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OpenStudy (mrhoola):
so maybe we have to integrate
. have you learned intgration
OpenStudy (anonymous):
no i haven't
OpenStudy (mrhoola):
ok i got it !!! woohoo
OpenStudy (anonymous):
how?
OpenStudy (mrhoola):
average distance = change in postition divided by change in time= or equivalently = (160 cm -0 cm) divided by ( 4/7 sec - 0 sec) = 280
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