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 m. A squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is 1.0 m above the ground on the other tree.Then he jumps back to first tree, landing on it at a spot that is 1.7m above the ground Finally, he leaps back to the other tree, now landing at a spot that is 2.5 m above the ground. What is the magnitude of the squirrel's displacement?

Options

A. 1.3 m
B. 2.8 m
C. 2.5 m
D. 3.4 m

Answer

To find the magnitude of the squirrel's displacement after its jumps between the two trees, we can break down the problem into steps:

### Step 1: Identify the initial and final positions of the squirrel.
- The squirrel starts at the base of the first tree (let's call this Tree A) at a height of 0 m.
- After the first jump, the squirrel lands on the second tree (Tree B) at a height of 1.0 m.
- After the second jump, the squirrel returns to Tree A at a height of 1.7 m.
- Finally, the squirrel jumps back to Tree B at a height of 2.5 m.

### Step 2: Determine the coordinates of the initial and final positions.
- The initial position (at the base of Tree A) can be represented as \( (0, 0) \).
- The first jump takes the squirrel to Tree B at \( (1.3, 1.0) \).
- The second jump brings the squirrel back to Tree A at \( (0, 1.7) \).
- The final position after the last jump is at Tree B at \( (1.3, 2.5) \).

### Step 3: Calculate the displacement vector.
- The displacement vector can be calculated from the initial position \( (0, 0) \) to the final position \( (1.3, 2.5) \).
- The displacement vector \( \vec{D} \) is given by:
\[ \vec{D} = (x_f - x_i, y_f - y_i) = (1.3 - 0, 2.5 - 0) = (1.3, 2.5) \]

### Step 4: Calculate the magnitude of the displacement.
- The magnitude of the displacement \( D \) can be calculated using the Pythagorean theorem:
\[ D = \sqrt{(x_f - x_i)^2 + (y_f - y_i)^2} = \sqrt{(1.3)^2 + (2.5)^2} \]
- Calculate \( (1.3)^2 \) and \( (2.5)^2 \):
\[ (1.3)^2 = 1.69 \quad \text{and} \quad (2.5)^2 = 6.25 \]
- Now sum these values:
\[ D = \sqrt{1.69 + 6.25} = \sqrt{7.94} \]
- Finally, calculate the square root:
\[ D \approx 2.81 \, \text{m} \]

### Conclusion:
The magnitude of the squirrel's displacement is approximately **2.81 meters**.

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