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Chapter 3: Problem 55

A police officer is driving due north at a constant speed of 29 m/s relative to the ground when she notices a truck on an east鈥搘est highway ahead of her, driving west at high speed. She finds that the truck鈥檚 speed relative to her car is 48 m/s (about 110 mph). (a) Sketch the vector triangle that shows how the truck鈥檚 velocity relative to the ground is related to the police car鈥檚 velocity relative to the ground and to the truck鈥檚 velocity relative to the police car. The sketch need not be to scale, but the ve ocity vectors should be oriented correctly and bear the appropriate abels. (b) What is the truck鈥檚 speed, relative to the ground?

Short Answer

Expert verified
The truck's speed relative to the ground is approximately 56.1 m/s.

Step by step solution

01

Understand the Problem

We are given the velocities of a police car and a truck, relative to each other and to the ground. The police officer is driving north and the truck is driving west. We need to find the truck's velocity relative to the ground.
02

Define the Velocity Vectors

Let's denote the police car's velocity relative to the ground as \(\overrightarrow{v_{pc,g}} = (0, 29)\) m/s (since it's going north, we consider it in the positive y-direction) and the truck's velocity relative to the police car as \(\overrightarrow{v_{t,pc}} = (v_{tx}, v_{ty})\), where \(v_{tx}\) is the truck's velocity in the east-west direction (negative, since it's going west), and \(v_{ty} = 0\) because the truck has no relative north-south movement compared to the police car.
03

Apply Relative Velocity Equation

We have \(\overrightarrow{v_{t,pc}} = \overrightarrow{v_{t,g}} - \overrightarrow{v_{pc,g}}\). Given \(\overrightarrow{v_{t,pc}} = (-48, 0)\) m/s, since the truck moves left (or west) relative to the police car. Substitute this into the equation: \((-48, 0) = (v_{tx}, v_{ty}) - (0, 29)\).
04

Solve the Velocity Components

From \((-48, 0) = (v_{tx}, v_{ty}) - (0, 29)\), we have two equations: 1. \(-48 = v_{tx}\) 2. \(v_{ty} = 29\) Thus, the truck's velocity relative to the ground is \(\overrightarrow{v_{t,g}} = (-48, 29)\) m/s.
05

Calculate the Truck's Speed Relative to the Ground

The speed of the truck relative to the ground is the magnitude of its velocity vector: \[|\overrightarrow{v_{t,g}}| = \sqrt{(-48)^2 + (29)^2} = \sqrt{2304 + 841} = \sqrt{3145} \approx 56.1\] m/s. Thus, the truck's speed relative to the ground is about 56.1 m/s.

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Key Concepts

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

Vector Triangle
A vector triangle is an essential tool in understanding relative motion between objects. In this exercise, the vector triangle helps visualize the velocities of the police car and the truck. Each side of the triangle represents a velocity vector.
  • The first vector points north along the y-axis, representing the police car's velocity of 29 m/s.
  • The second vector points west, depicting the truck鈥檚 velocity relative to the police car as 48 m/s.
  • The third vector is the truck's velocity relative to the ground, connecting the tips of the first two vectors.
This vector triangle is critical as it graphically shows how these velocity vectors relate to each other, helping solve problems involving relative velocity.
Velocity Vectors
Velocity vectors describe both the speed and direction of an object鈥檚 movement. They can be added or subtracted using vector components. This problem involves three key velocity vectors:
  • he police car鈥檚 velocity relative to the ground, overrightarrow{v_{pc,g}} = (0, 29) m/s, moving north.
  • the truck鈥檚 velocity relative to the police car, overrightarrow{v_{t,pc}} = (-48, 0) m/s, moving west.
  • the truck鈥檚 velocity relative to the ground, overrightarrow{v_{t,g}} = (-48, 29) m/s.
These vectors can be represented in component form, making it easier to manage calculations, such as finding the resultant vector or determining speed using the Pythagorean theorem.
Magnitude of Velocity
The magnitude of a velocity vector represents the speed of the object. It is found by calculating the length of the vector using its components. This involves: First, square each component of the vector.Next, add these squared values together.Finally, take the square root of the sum to determine the magnitude.For the truck鈥檚 velocity relative to the ground, this comes out as:\[|\\overrightarrow{v_{t,g}}\\\| = \\sqrt{(-48)^2 + (29)^2} = \\sqrt{2304 + 841} = \\sqrt{3145} \\approx 56.1 \text{ m/s}\\]This magnitude tells us the truck鈥檚 speed relative to the ground, which helps us quantitatively understand its motion.
Relative Motion
Relative motion involves observing the motion of an object as seen from another moving object. It requires understanding how velocities add or subtract depending on direction and observer.In this scenario:
  • The police car moves north, and when observing the truck, its motion appears as 48 m/s westward.
  • To determine the truck鈥檚 absolute motion relative to the ground, we consider both the police car's and truck's movements.
By applying relative velocity concepts, we use the equation:\(\overrightarrow{v_{t,pc}} = \\overrightarrow{v_{t,g}} - \\overrightarrow{v_{pc,g}} \)This shows how understanding relative motion allows us to predict and calculate actual trajectories and speeds accurately, crucial in navigation and tracking.

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