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Chapter 3: Problem 1

SSM 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 frisky 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 the first tree, landing on it at a spot that is 1.7 m 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鈥檚 displacement?

Short Answer

Expert verified
The squirrel's displacement is approximately 2.82 meters.

Step by step solution

01

Understanding Displacement

Displacement is a vector quantity that refers to the change in position of an object. It is the straight-line distance from the starting point to the ending point, regardless of the path taken. In this problem, we need to find the direct line distance from the starting point on the first tree to the final position on the second tree.
02

Determine Starting and Ending Points

The squirrel starts at the base of the first tree and ends 2.5 meters above the ground on the second tree. Therefore, the initial position is (0, 0) and the final position is (1.3, 2.5), thanks to the horizontal separation between the trees and the vertical height at which the squirrel lands on the second tree.
03

Using the Distance Formula

To calculate the magnitude of the displacement, which is basically the straight-line distance between these two points, we use the distance formula derived from the Pythagorean theorem:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]where \( (x_1, y_1) = (0, 0) \),and \( (x_2, y_2) = (1.3, 2.5) \).
04

Substitute Values and Calculate

Substitute the values into the distance formula:\[d = \sqrt{(1.3 - 0)^2 + (2.5 - 0)^2} = \sqrt{1.3^2 + 2.5^2} = \sqrt{1.69 + 6.25} = \sqrt{7.94}\]After calculating, we find:\[d \approx 2.82 \text{ meters}\]
05

Conclusion

The magnitude of the squirrel鈥檚 displacement, which is the direct distance from the starting point at the base of the first tree to the final position on the second tree, is approximately 2.82 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Displacement
The concept of vector displacement is essential in understanding how objects move from one point to another. Displacement is a vector quantity, meaning it has both magnitude and direction. In simpler terms, vector displacement is not just about the distance traveled but the "straight-line鈥� distance from the starting point to the ending point.

Imagine you're walking from your house to the park around a building. Even if you take a winding path, the vector displacement is the shortest path from your house directly to the park.
  • The squirrel, in our problem, made multiple jumps. Regardless of these jumps, what matters for vector displacement is the direct path from where it started to its final landing spot.
  • Vector displacement helps us determine how far, and in which direction, an object is from its original position, ignoring its path.
Recognizing this can simplify complex movements into a simple straight-line distance evaluation.
Distance Formula
The distance formula is a tool used to compute the straight-line distance between two points in a plane. This formula is especially useful when the movement involves a coordinate system, such as navigating across horizontal and vertical axes like grids or graphs.

Mathematically, the distance between the points \(x_1, y_1\) and \(x_2, y_2\) is calculated as: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
  • In our problem, the squirrel moves from \(0, 0\) (ground level on the first tree) to \(1.3, 2.5\) (the final jump spot on the second tree).
  • This formula is derived from the principle used in the Pythagorean theorem, which allows us to determine the 'straight-line' or Euclidean distance despite the path taken by the object.
Using the distance formula simplifies finding how far apart two positions are while accounting for both horizontal and vertical motion.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry that relates the lengths of the sides of a right-angled triangle. This theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Expressed mathematically as: \[c^2 = a^2 + b^2\]
  • In the context of the squirrel's jumps, we can create a right triangle where the horizontal distance between trees forms one leg and the vertical difference in landing heights forms the other.
  • The distance across the direct path, or displacement, is the hypotenuse.
By substituting the known values, we solve for this hypotenuse, using the Pythagorean theorem conceptually embedded in the distance formula.
Squirrel Displacement Calculation
In solving the problem of the squirrel's displacement, think of it as finding how far the squirrel is from its starting point directly to its final landing spot. This displacement involves accounting for both horizontal and vertical changes in position.

The process involves several straightforward steps:
  • Identify the initial and final positions where the squirrel lands \(0, 0\) to \(1.3, 2.5\).
  • Use the distance formula, which utilizes the Pythagorean theorem, to find the straight-line distance.
  • Calculate it step by step to ensure understanding, substituting coordinates into the formula correctly.
Ultimately, the calculated displacement provides a concise answer to how far the squirrel is from its starting point, ignoring the zigzag path of its multiple jumps. This answers the problem question directly, providing a handy application of vector displacement and basic geometric principles.

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Most popular questions from this chapter

A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in Figure 3.10 and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?

In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 35.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m. Ignore air resistance. (a) The first balloon launched by the Jackson team hits Walton dorm 2.0 s after launch, striking it halfway be- tween the ground and the roof. Find the direction of the balloon鈥檚 initial velocity. Give your answer as an angle measured above the horizontal. (b) A second balloon launched at the same angle hits the edge of Walton鈥檚 roof. Find the initial speed of this second balloon.

A dolphin leaps out of the water at an angle of \(35^{\circ}\) above the horizontal. The horizontal component of the dolphin's velocity is 7.7 \({m} / {s}\) . Find the magnitude of the vertical component of the velocity.

Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.
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