anonymous
  • anonymous
A weight is attached to one end of a 33 foot rope which passes over a pulley 18 feet above the ground. the other end is attached to a truck at a point 3 feet above the ground. if the truck moves away at a rate of 2 feet per second, how fast is the weight rising when the truck is 8 feet from the spot directly under the pulley?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
perl
  • perl
|dw:1349004422200:dw|
goformit100
  • goformit100
|dw:1349004500723:dw|
anonymous
  • anonymous
I need a solution :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

perl
  • perl
which direction is the trick moving, it is not clear
anonymous
  • anonymous
|dw:1349004602933:dw|
goformit100
  • goformit100
|dw:1349004739213:dw|
anonymous
  • anonymous
|dw:1349004667017:dw|
anonymous
  • anonymous
|dw:1349004778629:dw|
anonymous
  • anonymous
dx/dt =2 feet per second y is constant..... now, can u find dh/dt
anonymous
  • anonymous
how?
anonymous
  • anonymous
|dw:1349004927462:dw|
anonymous
  • anonymous
Since y is constant dy/dt=0 right?
anonymous
  • anonymous
yes. then?
anonymous
  • anonymous
|dw:1349005061647:dw|
anonymous
  • anonymous
Now, given x=8
anonymous
  • anonymous
y=18-3=15
anonymous
  • anonymous
So, |dw:1349005186713:dw|
anonymous
  • anonymous
right?
anonymous
  • anonymous
@sauravshakya the answer is 16/ 17
goformit100
  • goformit100
Use of calculus is good
anonymous
  • anonymous
|dw:1349005396208:dw|
anonymous
  • anonymous
Now we need to find dl/dt
anonymous
  • anonymous
how to fnd it @sauravshakya
perl
  • perl
you need a better diagram
perl
  • perl
A weight is attached to one end of a 33 foot rope which passes over a pulley 18 feet above the ground. the other end is attached to a truck at a point 3 feet above the ground. if the truck moves away at a rate of 2 feet per second, how fast is the weight rising when the truck is 8 feet from the spot directly under the pulley?
anonymous
  • anonymous
|dw:1349005616181:dw||dw:1349005671316:dw|
anonymous
  • anonymous
Did I did some algebra wrong or what???????
anonymous
  • anonymous
hmm i dont know.. the answer should be 16/ 17
perl
  • perl
|dw:1349005592570:dw|
perl
  • perl
x is the distance the weight has been raised, and y is the horizontal distance, where the rope is attached, to the vertical line passing through the pulley.
goformit100
  • goformit100
its ok
anonymous
  • anonymous
@bii17 u sure the answer is 16/17
goformit100
  • goformit100
its correct
anonymous
  • anonymous
@sauravshakya yes im sure. that is the answer given in the book
anonymous
  • anonymous
But I got 16/15 @goformit100
goformit100
  • goformit100
i did the calculation @sauravshakya is correct now
anonymous
  • anonymous
@ganeshie8 if u r free PLZ see this
goformit100
  • goformit100
@Callisto check it
goformit100
  • goformit100
plz
anonymous
  • anonymous
@hartnn can U plz check it
perl
  • perl
oh i will do it again
perl
  • perl
|dw:1349007585718:dw|
perl
  • perl
15 ^2 + x^2 = (33- y)^2
perl
  • perl
y is the height that the weight rises from the ground. , x is the distance the truck has moved to the left
hartnn
  • hartnn
|dw:1349007886322:dw| @sauravshakya
perl
  • perl
did you get 16/17?
anonymous
  • anonymous
Oh ya....... @hartnn
hartnn
  • hartnn
that still doesn't give answer.....
anonymous
  • anonymous
Yep
perl
  • perl
15 ^2 + x^2 = (33- y)^2 2x *dx/dt = 2 ( 33 - y) * (-dy/dt) x * dx/dt = (33-y) * (-dy/dt) x/ (33-y) * dx/dt = -dy/dt , this works
perl
  • perl
8 / ( 33 - 16 ) * 2 = 16/17 , x = 8 , that is given, y = 16 by plugging in 15^2 + x^2 = (33-y)^2
perl
  • perl
the only weird part is the negative sign
anonymous
  • anonymous
@hartnn that gives the answer
perl
  • perl
oh nevermind, it should be plus
perl
  • perl
15^2 + x^2 = (33+y)^2
anonymous
  • anonymous
|dw:1349008213306:dw|
perl
  • perl
|dw:1349008265370:dw|
anonymous
  • anonymous
So,|dw:1349008256578:dw|
hartnn
  • hartnn
yes it does, u did it @sauravshakya ! :)
perl
  • perl
yes that is simpler. the trick is to see that y is constant
perl
  • perl
|dw:1349008430327:dw|
perl
  • perl
y = 15,
perl
  • perl
wait, dh/dt is the rate at which the rope is being pulled
perl
  • perl
so dh/dt = the rate at which the weight is moving up ? interesting

Looking for something else?

Not the answer you are looking for? Search for more explanations.