# Boy Scout Problem Homework: Solve Kinematics Equations

• abhikesbhat
In summary, the boy scout raises his food pack by throwing a rope over a tree limb and walking away with constant velocity vboy. The speed of the food pack, v, can be expressed as x(x2 + h2)–1/2 vboy, where x is the distance he has walked away from the vertical rope. The acceleration of the food pack, a, is given by h2(x2 + h2)–3/2 v2boy. Shortly after the boy leaves the point under the pack (x = 0), the acceleration and velocity v have values of 0. As the distance x continues to increase, the pack's velocity and acceleration approach 0

## Homework Statement

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 vboy, 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(x2 + h2)–1/2 vboy where x is the distance he has
walked away from the vertical rope. (b) Show that the acceleration a of the food pack is
h2(x2 + h2)–3/2 v2
boy. (c) What values do the acceleration 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?

## Homework Equations

All the kinematics equations and Pythagorean Theorem.

## The Attempt at a Solution

I have been working at this for awhile now and can't seem to get it. I know that x=vboyt. I know that there is some trig here along the lines, but I can't get how the food box moves. Ok if the boy moves x then l will grow also, but at what rate? However much l grows is how high up the box goes. I can't seem to figure it out.

http://img443.imageshack.us/img443/9238/physicsquestionku9.jpg [Broken]

Last edited by a moderator:
Since you are interested in dy/dt (velocity of the pack) then express y in terms of x.

Pythagoras was good for more than building pyramids.

When you say y, are u talking about h?

abhikesbhat said:
When you say y, are u talking about h?

Y is a direction.

h is a height above the boy's head. That makes h a scalar constant doesn't it?

Using the Pythagorean Theorem express the hypotenuse Hy as a function of x.

You know as in

Hy2 = x2 + h2

Don't changes in the Hy represent 1:1 changes in y?

So the first question resolves into

dy/dt = dHy/dt

And Hy is a function of t because x is a function of t.

Ok so Hy=sqrt(x^2+h^2). What does that do? I'm sorry I clearly don't understand this concept... Let me review it and I'll come back to it. Also I haven't learned too much calculus, only knowing basics, so if there is an algebraic approach to this I might I understand it better. Sorry for your time LowlyPion.

abhikesbhat said:
Ok so Hy=sqrt(x^2+h^2). What does that do? I'm sorry I clearly don't understand this concept... Let me review it and I'll come back to it. Also I haven't learned too much calculus, only knowing basics, so if there is an algebraic approach to this I might I understand it better. Sorry for your time LowlyPion.

Just think about it conceptually. Boy runs with a velocity (dx/dt).

How does Hy change with x ? You have that equation.

So maybe take the derivative of Hy with respect to x? What does that give you? Since the rate of change of Hy = rate of change of y ... as it must since the rope doesn't change length ...

## 1. What is the Boy Scout Problem in regards to kinematics equations?

The Boy Scout Problem is a classic physics problem that involves a boy scout walking along a path at different speeds and directions, and the goal is to determine their displacement and final position.

## 2. What are the kinematics equations used to solve the Boy Scout Problem?

The three primary kinematics equations used to solve the Boy Scout Problem are the equations for displacement, velocity, and acceleration. These are:

- Displacement: d = xf - xi

- Velocity: v = (xf - xi) / t

- Acceleration: a = (vf - vi) / t

## 3. How do you apply the kinematics equations to solve the Boy Scout Problem?

First, you need to identify the initial and final positions and times of the boy scout's journey. Then, plug these values into the appropriate kinematics equations to solve for displacement, velocity, and acceleration. Make sure to pay attention to the units of measurement and use the correct formula for each variable.

## 4. Can the Boy Scout Problem be solved using only two of the kinematics equations?

Yes, it is possible to solve the Boy Scout Problem using only two of the kinematics equations. However, it is recommended to use all three equations to check for consistency and accuracy in the solution.

## 5. Are there any common mistakes to avoid when solving the Boy Scout Problem using kinematics equations?

One common mistake is to mix up the initial and final positions or times when plugging them into the equations. It is important to double check these values before solving. Also, make sure to use the correct formula for each variable, as using the wrong formula can lead to incorrect solutions.