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Chapter 2: Problem 8

Two cars are moving on the same straight-line road. Car \(\mathrm{A}\) moves to the right at velocity \(80 \mathrm{km} \mathrm{h}^{-1}\) and car \(\mathrm{B}\) moves at \(50 \mathrm{km} \mathrm{h}^{-1}\) to the left. Both velocities are measured by an observer at rest on the road. (a) Find the relative velocity of car \(\mathrm{B}\) with respect to car \(\mathrm{A}\) . (b) Find the relative velocity of car \(\mathrm{A}\) with respect to car \(\mathrm{B}\).

Short Answer

Expert verified
a) -130 km/h, b) 130 km/h

Step by step solution

01

Understanding Relative Velocity

Relative velocity is the velocity of one object as observed from another moving object. If two objects are moving towards each other, their relative velocity is the sum of their individual velocities.
02

Define Given Velocities

Car A has a velocity of 80 km/h to the right. Car B has a velocity of 50 km/h to the left. Note that leftward velocity is negative, so we have: \[v_{A} = +80 \, km/h \] \[v_{B} = -50 \, km/h \]
03

Calculate Relative Velocity of Car B with Respect to Car A

Use the formula for relative velocity: \[v_{B \, relative \, to \, A} = v_{B} - v_{A} \] Substitute the velocities: \[v_{B \, relative \, to \, A} = -50 \, km/h - 80 \, km/h \] Simplify: \[v_{B \, relative \, to \, A} = -130 \, km/h \] So, car B moves left at 130 km/h relative to car A.
04

Calculate Relative Velocity of Car A with Respect to Car B

Similarly, use the formula for relative velocity: \[v_{A \, relative \, to \, B} = v_{A} - v_{B} \] Substitute the velocities: \[v_{A \, relative \, to \, B} = 80 \, km/h - (-50 \, km/h) \] Simplify: \[v_{A \, relative \, to \, B} = 80 \, km/h + 50 \, km/h \] \[v_{A \, relative \, to \, B} = 130 \, km/h \] So, car A moves right at 130 km/h relative to car B.

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Key Concepts

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

Relative Motion
Relative motion is simply the movement of an object as observed from another moving object. Imagine sitting in a moving car and looking at another car passing by. The way you perceive the speed and direction of the passing car isn't just about how fast the other car is moving, but also about how fast your car is moving. This idea helps us understand the concept of relative motion.

In our exercise, we have two cars - Car A and Car B - traveling in opposite directions. Car A moves to the right at a speed of 80 km/h, and Car B moves to the left at a speed of 50 km/h. By understanding relative motion, we can figure out how fast one car appears to move when viewed from the other car.
Velocity Subtraction
Velocity subtraction is used to find the relative velocity between two moving objects. The relative velocity tells us how fast one object is moving compared to the other.

To calculate the relative velocity of Car B with respect to Car A, we use the formula: \( v_{B \, relative \, to \, A} = v_{B} - v_{A} \).

By substituting the given velocities: \( v_{B \, relative \, to \, A} = -50 \, km/h - 80 \, km/h \) (since Car B's leftward velocity is negative), we get \( v_{B \, relative \, to \, A} = -130 \, km/h \). This means Car B appears to move to the left at 130 km/h relative to Car A.

Similarly, the relative velocity of Car A with respect to Car B is given by: \( v_{A \, relative \, to \, B} = v_{A} - v_{B} \) and by substituting the velocities: \( v_{A \, relative \, to \, B} = 80 \, km/h - (-50 \, km/h) = 130 \, km/h \). This tells us that Car A appears to move to the right at 130 km/h relative to Car B.
Direction of Motion
Understanding the direction of motion is crucial when dealing with relative velocity. In our example, Car A moves to the right and Car B moves to the left. It's important to assign a positive or negative sign to these directions to make calculations easier.

Typically, we assign positive values to the rightward direction and negative values to the leftward direction. So, Car A's velocity is +80 km/h, and Car B's velocity is -50 km/h.

By keeping track of these directions, we can correctly apply the velocity subtraction formula. When you subtract a negative velocity, it effectively adds the magnitude. This makes calculating relative velocities straightforward and helps in visualizing how one object appears to move relative to the other.

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Most popular questions from this chapter

Show by applying Newton's law of gravitation and the second law of mechanics that a satellite (or planet) in a circular orbit of radius \(R\) around the earth (or the sun) has a period (i.e. time to complete one revolution) given by $$T^{2}=\frac{4 \pi^{2} R^{3}}{G M}$$ where \(M\) is the mass of the attracting body (earth or sun). This is Kepler's third law.

If you jump from a height of \(1.0 \mathrm{m}\) from the surface of the earth, the earth actually moves up a bit as you fall. (a) Explain why. (b) Estimate the distance the earth moves, listing any assumptions you make. (c) Would the earth move more, less or the same if a heavier person jumps?

A ball is thrown upward from the edge of a cliff with velocity \(20.0 \mathrm{m} \mathrm{s}^{-1} .\) It reaches the bottom of the cliff 6.0 s later. (a) How high is the cliff? (b) With what speed does the ball hit the ground?

Two cars of masses \(1200 \mathrm{kg}\) and \(1400 \mathrm{kg}\) collide head-on and stick to each other. The cars are coming at each other from opposite directions with speeds of \(8.0 \mathrm{m} \mathrm{s}^{-1}\) and \(6 \mathrm{m} \mathrm{s}^{-1}\), respectively. With what velocity does the wreck move away from the scene of the accident?

A rocket in space where gravity is negligible has a mass (including fuel) of \(5000 \mathrm{kg}\). If it is desired to give the rocket an average acceleration of \(15.0 \mathrm{m} \mathrm{s}^{-2}\) during the first second of firing the engine and the gases leave the rocket at a speed of \(1500 \mathrm{m} \mathrm{s}^{-1}\) (relative to the rocket), how much fuel must be burned in that second?
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