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How long does it take an automobile traveling in the left lane at \(60.0 \mathrm{km} / \mathrm{h}\) to pull alongside a car traveling in the same direction in the right lane at \(40.0 \mathrm{km} / \mathrm{h}\) if the cars' front bumpers are initially 100 m apart?

Short Answer

Expert verified
It takes 18 seconds for the automobile to pull alongside the other car.

Step by step solution

01

Convert Speed from km/hr to m/s

Firstly, speeds need to be in the same units as the distance - in this case meters. The conversion factor from km/hr to m/s is 1km/hr = 0.277778 m/s. So, the speed of the fast car is \(60.0 \mathrm{km}/\mathrm{hr} * 0.277778 \mathrm{m}/\mathrm{s}/\mathrm{km}/\mathrm{hr} = 16.67 \mathrm{m}/\mathrm{s}\) and the speed of the slower car is \(40.0 \mathrm{km}/\mathrm{hr} * 0.277778 \mathrm{m}/\mathrm{s}/\mathrm{km}/\mathrm{hr}= 11.11 \mathrm{m}/\mathrm{s}\)
02

Calculate Relative Speed

In this scenario, the faster car is trying to catch up with the slower car, so we are interested in the relative speed (i.e. the speed difference). This is calculated as \(16.67 \mathrm{m}/\mathrm{s} - 11.11 \mathrm{m}/\mathrm{s} = 5.56 \mathrm{m}/\mathrm{s}\)
03

Calculate Time

We use the formula for time, which is distance divided by speed. In this case, the distance is the initial separation (100 m) and the speed is the relative speed (5.56 m/s). So the time is \(100 \mathrm{m} / 5.56\mathrm{m}/\mathrm{s} = 18 \mathrm{s}\)

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Key Concepts

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

Understanding Kinematics
Kinematics is a branch of classical mechanics that deals with the motion of objects without considering the forces that cause the motion. A foundational understanding of kinematics is crucial for various fields of physics and engineering. At its core, kinematics focuses on the trajectory of an object as it moves, which is determined by variables such as displacement, velocity, acceleration, and time.

When analyzing problems in kinematics, it's essential to understand that motion is relative. This means that the velocity of an object can vary depending on the observer's frame of reference. For example, if you're sitting in a train moving at a constant speed and another train on a parallel track is moving slower than yours, it will appear to be moving backward relative to your position. This effect is due to the concept of relative motion, which is a key topic in problems like the one stated in the exercise, where two vehicles are moving in the same direction at different speeds.

In solving kinematic problems, it's also vital to represent all physical quantities in consistent units, use clear diagrams to illustrate motion, and apply appropriate equations to connect the various kinematic variables.
Speed Conversion for Consistent Units
Speed conversion is essential in physics to ensure that all variables are expressed in the same units, allowing for accurate calculations and comparisons. For instance, in many problems, distance might be measured in meters (m), while speed could initially be given in kilometers per hour (km/h). Conversion between these units is crucial for working out problems accurately.

To convert km/h to m/s, which is a standard unit for speed in scientific calculations, you multiply the speed by a conversion factor. This factor is 0.277778 m/s for every km/h, as shown in the step-by-step solution provided with the exercise. Being comfortable with this conversion factor allows students to approach motion problems with ease and ensures calculation integrity.

In problems involving time, it's often useful to convert hours to seconds, or vice versa, as well. Remembering that 1 hour equals 3600 seconds is key for such conversions. Frequent practice with these conversions builds a student’s confidence in tackling a wide range of physics problems.

Conversion Tip:

Always double-check your unit conversions to avoid simple mistakes that could lead to incorrect solutions.
Calculating Relative Speed
Relative speed calculation is a crucial skill when dealing with scenarios where two or more objects are moving in relation to each other. Relative speed is essentially the speed of one object as observed from another moving object. In the context of our example exercise, it's the difference in speed between the two cars as one tries to overtake the other.

To calculate the relative speed, you subtract the speed of the slower object from the speed of the faster object when both are moving in the same direction. However, if the objects are moving in opposite directions, their relative speed is the sum of their individual speeds. This concept hinges on understanding that speed is a scalar quantity – it has magnitude but no direction.

The use of relative speed simplifies the problem, allowing us to treat the slower car as if it were stationary and focus on how quickly the faster car closes the distance between them. Remember, when objects move in the same direction, the relative speed is lower than when they move towards each other because they converge at a slower rate.

Key Point:

Always define the direction of motion and the frame of reference when dealing with relative speed to maintain a clear understanding of the problem at hand.

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Most popular questions from this chapter

The vector position of a particle varies in time according to the expression \(\mathbf{r}=\left(3.00 \hat{\mathbf{i}}-6.00 t^{2} \hat{\mathbf{j}}\right) \mathrm{m} .\) (a) Find expres- sions for the velocity and acceleration as functions of time. (b) Determine the particle's position and velocity at \(t=1.00 \mathrm{s}\)

Young David who slew Goliath experimented with slings before tackling the giant. He found that he could revolve a sling of length \(0.600 \mathrm{m}\) at the rate of 8.00 rev/s. If he increased the length to \(0.900 \mathrm{m},\) he could revolve the sling only 6.00 times per second. (a) Which rate of rotation gives the greater speed for the stone at the end of the sling? (b) What is the centripetal acceleration of the stone at 8.00 rev/s? (c) What is the centripetal acceleration at \(6.00 \mathrm{rev} / \mathrm{s} ?\)

\- As their booster rockets separate, Space Shuttle astronauts typically feel accelerations up to \(3 g,\) where \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) In their training, astronauts ride in a device where they experience such an acceleration as a centripetal acceleration. Specifically, the astronaut is fastened securely at the end of a mechanical arm that then turns at constant speed in a horizontal circle. Determine the rotation rate, in revolutions per second, required to give an astronaut a centripetal acceleration of \(3.00 g\) while in circular motion with radius \(9.45 \mathrm{m}\)

Two swimmers, Alan and Beth, start together at the same point on the bank of a wide stream that flows with a speed v. Both move at the same speed \(c(c>v),\) relative to the water. Alan swims downstream a distance \(L\) and then \(\mathrm{up}^{-}\) stream the same distance. Beth swims so that her motion relative to the Earth is perpendicular to the banks of the stream. She swims the distance \(L\) and then back the same distance, so that both swimmers return to the starting point. Which swimmer returns first? (Note: First guess the answer.)

A motorist drives south at \(20.0 \mathrm{m} / \mathrm{s}\) for \(3.00 \mathrm{min}\) then turns west and travels at \(25.0 \mathrm{m} / \mathrm{s}\) for \(2.00 \mathrm{min},\) and finally travels northwest at \(30.0 \mathrm{m} / \mathrm{s}\) for 1.00 min. For this 6.00 -min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive \(x\) axis point east.

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