/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 How long does it take an automob... [FREE SOLUTION] | 91Ó°ÊÓ

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How long does it take an automobile traveling in the left lane of a highway at \(60.0 \mathrm{~km} / \mathrm{h}\) to overtake (become even with) another car that is traveling in the right lane at \(40.0 \mathrm{~km} / \mathrm{h}\) when the cars' front bumpers are initially \(100 \mathrm{~m}\) apart?

Short Answer

Expert verified
The car in the left lane will take \(18.0 \mathrm{s}\) to overtake the car in the right lane.

Step by step solution

01

Convert speed from km/h to m/s

First, convert speeds from km/h to m/s for each car. This is done by multiplying the speed by \(1000\) (to convert kilometers to meters) and then divided by \(3600\) (to convert hours to seconds). For the left lane car: \(60.0 \mathrm{~km/h} = 60.0 * 1000 / 3600 = 16.67 \mathrm{~m/s}\) and for the right lane car: \(40.0 \mathrm{~km/h} = 40.0 * 1000 / 3600 = 11.11 \mathrm{~m/s}\).
02

Calculate relative speed

Relative speed is the speed of one object as observed from another. Here, think from the perspective of the faster car in the left lane. So, the relative speed of the car in the left lane with respect to the car in the right lane is: \(16.67 \mathrm{~m/s} - 11.11 \mathrm{~m/s} = 5.56 \mathrm{~m/s}\).
03

Calculate time to overtake

Time is calculated by dividing the initial separation distance by the relative speed. So, time \(t = 100 \mathrm{~m} / 5.56 \mathrm{~m/s} = 18.0 \mathrm{~s}\).

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Key Concepts

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

Relative Speed Calculation
When discussing the concept of relative speed in physics, we are examining the scenario where two or more objects are moving in a system, and we want to know how fast one object is moving from the perspective of another. What one needs to consider is the direction of the objects’ movements. If two objects are moving in the same direction, the relative speed is the difference between their speeds. Conversely, if they are moving toward each other, their relative speeds add up.

Take, for instance, the textbook problem regarding two cars on a highway. The car in the left lane is travelling faster than the car in the right lane. To calculate the time it takes for the faster car to overtake the slower car, we need to compute their relative speed. You think about it as if you're in the faster car: all you care about is how quickly you're catching up to the car ahead. In this case, since both cars are moving in the same direction, we subtract the speed of the slower car from the speed of the faster car. An important note is that the speeds must be in consistent units, which leads us to the necessity of understanding unit conversions in physics.
Unit Conversion in Physics
In physics, measurements might be given in a range of units, and performing accurate calculations often requires converting these units into a standard or commonly used set. This becomes incredibly important in problems involving motion, like the overtake scenario posed in the textbook question.

For the textbook problem, the speeds of the cars are given in kilometers per hour (km/h), but to proceed with the calculation, we need the speeds to be in meters per second (m/s), which is a more practical unit for such computations. The process involves multiplying by 1000 (since there are 1000 meters in a kilometer) and then dividing by 3600 (since there are 3600 seconds in an hour). This is an essential step because if the units are not consistent, the answer will be incorrect. Remember, misunderstanding or misapplying unit conversion can dramatically change the outcome of a physics problem.
Time Calculation in Motion
The final part to solving motion problems like our overtaking cars example is to calculate time. The formula universally used in physics is quite straightforward: time equals distance divided by speed. But here we make use of the relative speed we've talked about because we're interested in how quickly the distance between two moving objects closes or increases.

In the textbook problem, the initial separation distance was 100 meters. With the relative speed we calculated, determining how long it takes for the separation distance to reduce to zero, hence for one car to overtake the other, is a simple division. It's essential to provide the time in the appropriate units to match the scenario. In physics, the emphasis on correct unit usage is constant since it ensures the answer will be meaningful in the real world context of the problem.

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Most popular questions from this chapter

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). The first is thrown at an angle of \(70.0^{\circ}\) with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?

A rowboat crosses a river with a velocity of \(3.30 \mathrm{mi} / \mathrm{h}\) at an angle \(62.5^{\circ}\) north of west relative to the water. The river is \(0.505 \mathrm{mi}\) wide and carries an eastward current of \(1.25 \mathrm{mi} / \mathrm{h}\). How far upstream is the boat when it reaches the opposite shore?

A chinook (king) salmon (genus Oncorhynchus) can jump out of water with a speed of \(6.26 \mathrm{~m} / \mathrm{s}\). (See Problem \(4.9\), page 111 for an investigation of how the fish can leave the water at a higher speed than it can swim underwater.) If the salmon is in a stream with water speed equal to \(1.50 \mathrm{~m} / \mathrm{s}\), how high in the air can the fish jump if it leaves the water traveling vertically upwards relative to the Earth?

A girl delivering newspapers covers her route by traveling \(3.00\) blocks west, \(4.00\) blocks north, and then \(6.00\) blocks east. (a) What is her resultant displacement? (b) What is the total distance she travels?

A river has a steady speed of \(v_{1}\) A student swims upstream a distance \(d\) and back to the starting point. (a) If the student can swim at a speed of \(v\) in still water, how much time \(t_{\text {op }}\) does it take the student to swim upstream a distance \(d ?\) Express the answer in terms of \(d\). \(v_{\text {, }}\) and \(v_{s}\). (b) Using the same variables, how much time \(t_{\text {doun }}\) does it take to swim back downstream to the starting point? (c) Sum the answers found in parts (a) and (b) and show that the time \(t_{a}\) required for the whole trip can be written as $$t_{\mathrm{a}}=\frac{2 d / v}{\mathrm{I}-\mathrm{v}_{1}^{2} / v^{2}}$$ (d) How much time \(t_{b}\) does the trip take in still water? (e) Which is larger, \(l_{a}\) or \(t_{b}^{2}\) Is it always larger?

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