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Chapter 2: Problem 15

To qualify for the finals in a racing event, a race car must achieve an average speed of \(250 \mathrm{~km} / \mathrm{h}\) on a track with a total length of \(1600 \mathrm{~m}\). If a particular car covers the first half of the track at an average speed of \(230 \mathrm{~km} / \mathrm{h}\), what minimum average speed must it have in the second half of the event in order to qualify?

Short Answer

Expert verified
The race car must average a speed of at least 274.03 km/h in the second half of the track in order to qualify.

Step by step solution

01

Determine time spent on the first half of the track

First, convert the speeds from km/h to m/s because the length of the track is given in meters. After the conversion of 230 km/h, multiply half the length of the track (800m) by the time to get the time spent on the first half. \[ Time_{first} = \frac{Distance_{first}}{Speed_{first}} = \frac{800m}{63.89 m/s} = 12.53s \]
02

Determine the time remaining for the rest of the track

Next, calculate the total time it should take to complete the track at an average speed of 250km/h. Convert this speed to m/s and then divide the total track length by this speed to determine the total time. Subtract the time spent on the first half to get the time remaining for the second half. \[ Time_{total} = \frac{Distance_{total}}{Speed_{total}} = \frac{1600m}{69.44 m/s} = 23.04s \] \[ Time_{second} = Time_{total} - Time_{first} = 23.04s - 12.53s = 10.51s \]
03

Calculate the minimum average speed for the second half of the track

Divide the half track distance by the time remaining for the second half to get the minimum average speed required in m/s. Convert this back to km/h to give the final answer. \[ Speed_{second} = \frac{Distance_{second}}{Time_{second}} = \frac{800m}{10.51s} = 76.12 m/s \] \[ Speed_{second} = 76.12 m/s * 3.6 = 274.03 km/h \]

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Key Concepts

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

Physics Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the causes of this motion. It covers concepts such as displacement, velocity, acceleration, and time. Average speed is a foundational concept in kinematics, representing the total distance traveled divided by the total time taken to traverse that distance.

In our race car example, understanding kinematics is crucial for solving the problem of the car achieving a qualifying average speed. The car's motion is split into two intervals, each with its own average speed, and kinematics principles are used to relate these speeds with distances and times to find the required speed for the second half of the track.
Speed Conversion
Speed conversion is essential in physics problems where units must match to perform calculations correctly. Common conversions involve changing kilometers per hour (km/h) to meters per second (m/s) and vice versa. The conversion factor between these units is 3.6 (1 m/s = 3.6 km/h).

In the given problem, speed conversion is necessary because the car's average speed requirement and initial speed are given in km/h, but the track length is in meters. Converting the car's initial speed of 230 km/h to m/s allows for computing the time it takes to cover the first half of the race track, and likewise, the required speed for the second half is computed in m/s before being converted back to km/h for the final answer.
Average Speed Calculation
The average speed calculation involves dividing the total distance traveled by the total time taken. This calculation is straightforward when speed is constant, but when it varies, as in the race car problem, we must consider each segment individually.

To determine the car's minimum speed for the second half to achieve the overall average speed, we calculate the time spent on each half and the distance covered. After finding the time remaining for the second half, we calculate the required average speed by dividing the second half's length by this remaining time. This demonstrates an important aspect of average speed: it doesn't necessarily remain constant across different intervals of a journey, and varying it strategically can help in meeting specific time or distance goals.

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Most popular questions from this chapter

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\), It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

A student throws a set of keys vertically upward to his fraternity brother, who is in a window a distance \(h\) above. The brother's outstretched hand catches the keys on their way up a time \(t\) later. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? (Answers should be in terms of \(h, g\), and \(t .)\)

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at \(71.5 \mathrm{~m} / \mathrm{s}\). The driver of the Thunderbird realizes that she must make a pit stop, and she smoothly slows to a stop over a distance of \(250 \mathrm{~m}\). She spends \(5.00 \mathrm{~s}\) in the pit and then accelerates out, reaching her previous speed of \(71.5 \mathrm{~m} / \mathrm{s}\) after a distance of \(350 \mathrm{~m}\). At this point, how far has the Thunderbird fallen behind the Mercedes Benz. which has continued at a constant speed?

A truck on a straight road starts from rest and accelerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) until it reaches a speed of \(20 \mathrm{~m} / \mathrm{s}\). Then the truck travels for \(20 \mathrm{~s}\) at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional \(5.0 \mathrm{~s}\). (a) How long is the truck in motion? (b) What is the average velocity of the truck during the motion described?

Two boats start together and race across a \(60-\mathrm{km}\) -wide lake and back. Boat A goes across at \(60 \mathrm{~km} / \mathrm{h}\) and returns at \(60 \mathrm{~km} / \mathrm{h}\). Boat \(\mathrm{B}\) goes across at \(30 \mathrm{~km} / \mathrm{h}\), and its crew, realizing how far behind it is getting, returns at \(90 \mathrm{~km} / \mathrm{h}\). Turnaround times are negligible, and the boat that completes the round trip first wins. (a) Which boat wins and by how much? (Or is it a tie?) (b) What is the average velocity of the winning boat?
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