/*! 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 34 A race driver has made a pit sto... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is \(6.0 \mathrm{m} / \mathrm{s}^{2}\); after \(4.0 \mathrm{s}\) he cnters the main specdway. At the same instant, another car on the speedway and traveling at a constant velocity of \(70.0 \mathrm{m} / \mathrm{s}\) overtakes and passes the entering car. The cntering car maintains its acceleration. How much time is required for the entering car to catch the other car?

Short Answer

Expert verified
The entering car catches the other car in approximately 14.2 seconds.

Step by step solution

01

Identify Variables and Conditions

Given the problem, we have two cars: - **Entering Car**: Starts from rest with an acceleration of \(a = 6.0 \ m/s^2\) and travels for \(t = 4.0 \ s\) while accelerating.- **Other Car**: Moves at a constant velocity \(v = 70.0 \ m/s\).We need to find the time taken for the entering car to catch up with the other car on the speedway.
02

Calculate Distance Covered by Entering Car till Speedway

The entering car accelerates for \(4 \ seconds\) in the pit area before entering the speedway. The distance \(s\) it covers is given by the equation of motion: \[ s = ut + \frac{1}{2} a t^2 \]where initial velocity \(u = 0\), acceleration \(a = 6 \ m/s^2\), and time \(t = 4 \ s\):\[ s = 0 + \frac{1}{2} \times 6 \times (4)^2 \]\[ s = 48 \ m \]
03

Determine Initial Speed of Entering Car at Speedway

The speed \(v_i\) of the entering car as it enters the speedway can be found using the formula:\[ v_i = u + at \]where \(u = 0\), \(a = 6 \ m/s^2\), and \(t = 4 \ s\):\[ v_i = 0 + 6 \times 4 = 24 \ m/s \]
04

Set Up Relative Motion Equation

At the moment the entering car enters the speedway, it must catch the other car which starts at that point a distance ahead. As both cars start at the same location after \(4 \ seconds\), their relative separation increases only due to differing speeds.- **Distance of Entering Car** from its starting point at any time \(t\) is:\[ s_1(t) = 48 + 24t + \frac{1}{2} \times 6 \times t^2 = 48 + 24t + 3t^2 \]- **Distance of Other Car** from its starting point at any time \(t\) is:\[ s_2(t) = 70t \]To catch the other car, their distances should be the same:\[ 48 + 24t + 3t^2 = 70t \]
05

Solve the Quadratic Equation

Rearrange the equation from Step 4 to form:\[ 3t^2 + 24t + 48 = 70t \]\[ 3t^2 - 46t + 48 = 0 \]Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = 3\), \(b = -46\), \(c = 48\):\[ t = \frac{46 \pm \sqrt{(-46)^2 - 4 \times 3 \times 48}}{2 \times 3} \]\[ t = \frac{46 \pm \sqrt{2116 - 576}}{6} \]\[ t = \frac{46 \pm \sqrt{1540}}{6} \]\[ t = \frac{46 \pm 39.24}{6} \]The valid solution is \( t = \frac{46 + 39.24}{6} = 14.2 \ s \) approximately, as time cannot be negative.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Relative Motion
In the world of kinematics, understanding relative motion is key. It involves analyzing how objects move in relation to one another. In this exercise, we deal with two cars: one accelerating from rest and the other moving at a constant speed. The main idea is to see how fast the first car needs to move to catch up with the second car.

To solve this, we consider the following aspects:
  • The relative speed between the two cars.
  • The distances each car travels in a given time.
  • The point at which the faster car, still accelerating, catches the slower, steady-speed car.

The relative motion equation sets the stage, showing how the difference in speeds influences their positions over time. The concept helps us model real-world problems like this one, where predicting motions relative to other objects is necessary.

In this problem, both cars begin at the same reference point after 4 seconds of the entering car accelerating. Here, relative motion isn't just about one car catching another. It demonstrates how motion depends on a reference frame, allowing us to understand distances and time intervals effectively. Remember, relative motion takes into account how one observer perceives the speed of one object against another in their frame.
Why Quadratic Equations Matter in Kinematics
Quadratic equations frequently pop up in kinematic problems, particularly those involving motion with constant acceleration. For those unfamiliar, a quadratic equation looks like this: \[ ax^2 + bx + c = 0 \]. It can have two possible solutions corresponding to different scenarios in motion problems.

In this exercise, a quadratic equation is formed when finding the time for the entering car to catch the other car. This arises when equating their distances over time, with each component having a different degree: constant, linear, and squared. Each term in the equation has significance:
  • The constant term reflects initial conditions or positions.
  • The linear term corresponds to velocity factors.
  • The squared term signifies acceleration.

Solving these equations typically involves using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. This approach identifies the points in time where conditions—such as meeting points of cars—occur.

The reason quadratic equations are vital in kinematics is their ability to manage scenarios where simple linear calculations fail. They handle cases of rapid change over time, like acceleration. Understanding how to deploy and solve quadratic equations is crucial as they often underpin the dynamics of moving objects.
Grasping Acceleration and Its Impacts
Acceleration is the rate at which an object changes its velocity. It's a central part of kinematics and crucial for understanding how quickly an object can achieve a certain speed. In our exercise, the entering car's acceleration determines its ability to catch the other car. With an acceleration of \(6 \, \text{m/s}^2\), the entering car speeds up from rest.

Key points to understanding acceleration include:
  • It's a vector, meaning it has both magnitude and direction.
  • It results in changes in speed and/or direction.
  • Most importantly, it defines how fast velocity changes over time.

When assessed with velocity, time, and distance, acceleration tells us the full story of motion over time. During the 4 seconds before entering the main speedway, the car's acceleration gives it the speed needed to eventually catch the steady-speed car. Understanding this gives insight into how long it takes for one moving object to match or surpass another's trajectory.

It's crucial to note: without acceleration, the car couldn't overcome the initial speed disadvantage relative to the other vehicle. Thus, comprehending acceleration's role is fundamental to solving real-world problems in physics, making it a powerful concept in kinematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Due to continental drift, the North American and European continents are drifting apart at an average speed of about \(3 \mathrm{cm}\) per year. At this speed, how long (in years) will it take for them to drift apart by another \(1500 \mathrm{m}\) (a little less than a mile)?

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board), assuming that the dowmward direction is chosen as the negative direction./ (b) What is the highest point he reaches above the water?

A hot-air balloon is rising straight up at a constant speed of \(7.0 \mathrm{m} / \mathrm{s} .\) When the balloon is \(12.0 \mathrm{m}\) above the ground, a gun fires a pellet straight up from ground level with an initial speed of \(30.0 \mathrm{m} / \mathrm{s}\). Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground are these places?

A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of \(8.0 \mathrm{s}\). What is its initial velocity? Neglect air resistance.

The three-toed sloth is the slowest-moving land mammal. On the ground, the sloth moves at an average speed of \(0.037 \mathrm{m} / \mathrm{s}\), considerably slower than the giant tortoise, which walks at \(0.076 \mathrm{m} / \mathrm{s}\). After 12 minutes of walking, how much further would the tortoise have gone relative to the sloth?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.