To protect his food from hungry bears, a boy scout raises...

To protect his food from hungry bears, a boy scout raises his food pack with a rope that is thrown over a tree limb at height $h$ above his hands. He walks away from the vertical rope with constant velocity $v$ boy, holding the free end of the rope in his hands (Fig. P2.71). (a) Show that the speed $v$ of the food pack is given by $x\left(x^{2}+h^{2}\right)^{-1 / 2} v_{\text { boy where } x}$ is the distance he has walked away from the vertical rope. (b) Show that the acceleration $a$ of the food pack is $h^{2}\left(x^{2}+h^{2}\right)^{-3 / 2} v_{\text { boy }}^{2}$ . (c) What values do the acceleration $a$ and velocity $v$ have shortly after he leaves the point under the pack $(x=0) ?$ (d) What values do the pack's velocity and acceleration approach as the distance $x$ continues to increase?

Transcript

00:01 So in this question, we've got a tree branch.

00:04 We've got a boy, and he's got a rope that goes over it, and it's got his food.

00:13 And the distance above his hands to the tree branch is h.

00:18 That's called the distance of the rope that he has is l, and the distance between him and his food is x.

00:25 And let's say that the distance of his food above his hands is y.

00:29 Then y is going to be l minus h because the total length of the rope is going to be 2h because when he was standing next to the food the food was by his hands so uh l minus h is going to be the distance up that it's gone but l is just the square root of h squared plus x squared so this is the expression for y so now the velocity of the food is going to be d by by d t of root h squared plus x squared minus h.

01:10 The h is a constant so it doesn't change.

01:14 And the derivative of the rest of it, we take the derivative with respect to x and then multiply by v boy, and that's going to be x v boy over root h squared plus x squared, which is what we wanted.

01:29 And now we want to have the acceleration and the acceleration is going to be dv by d t.

01:38 And so this is going to be, so v -boy is a constant, so we can take it out the front, and then this is going to be d -by -d -t of x, h -squared plus x squared to the minus a half.

01:54 And d -by -d -t of x is v -boy.

01:57 So we're going to do the product rule.

02:01 So we've got v -boy times h -squared plus x squared to the minus a half, and then we get minus x squared, because we're taking the derivative of this part now.

02:18 So we get a 2x that hits this x and becomes a 2x squared, but then we get a minus a half.

02:23 So we get minus x squared, h squared plus x squared to the minus three halves.

02:34 So this overall is v -boy squared over root h squared plus x squared.

02:48 Minus x squared over h squared plus x squared root h squared plus x squared um and now if we multiply by h squared plus x squared we're going to get that this is h squared plus x squared v boy squared minus x squared divided by h squared plus x squared to the three halves and, oh, sorry, and we get a v -boy in here as well...

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