Chapter 3, Problem 1P

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by $1.3 \text{m}.$ A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is $1.0 \text{m}$ above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is $1.7 \text{m}$ above the ground. Finally, he leaps back to the other tree, now landing at a spot that is $2.5 \text{m}$ above the ground. What is the magnitude of the squirrel’s displacement?

To determine

The displacement of the squirrel.

Answer to Problem 1P

  $2.8\text{m}$

Explanation of Solution

Given:

The side of the trunks of two parallel trees is parted by 1.3 m. A squirrel makes 3 jumps.

Formula used:

Using Pythagoras for triangle ABC,

  $AC=\sqrt{A{B}^2+B{C}^2}$

Calculation:

Initially, the squirrel jumps from the foot of one tree to another tree spots 1m above the ground.

Then squirrel hurdles back to the first tree $1.7\text{m}$ above the ground.

Now again squirrel leaps to another tree at a spot $2.5\text{m}$ above the ground.

So, the total displacement of the squirrel is the shortest distance between the initial and final positions.

  

The figure shows the initial position of the squirrel at A and the final position of a squirrel is C.

So, the shortest distance between A and C is given as,

  $\begin{array}{l}AC=\sqrt{A{B}^2+B{C}^2}\\ AC=\sqrt{{\left(\text{1}\text{.3m}\right)}^2+{\left(\text{2}\text{.5m}\right)}^2}\\ AC=\sqrt{3.\text{8m}}\\ AC=2.81\text{m}\\ AC=2.8\text{m}\end{array}$

Conclusion:

So, the magnitude of the displacement of the squirrel is $2.8\text{m}$ .