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Review Interactive LearningWare 2.2 at in preparation for this problem. A race driver has made a pit stop to refuel. After refueling, he leaves the pit area with an acceleration whose magnitude is \(6.0 \mathrm{~m} / \mathrm{s}^{2}\); after \(4.0 \mathrm{~s}\) he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant speed of \(70.0 \mathrm{~m} / \mathrm{s}\) overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?

Short Answer

Expert verified
The car takes 23.33 seconds to catch up.

Step by step solution

01

Understand the Problem

We have two cars: Car A starts from rest and accelerates at \(6.0 \mathrm{~m/s^2}\), while Car B moves at a constant speed of \(70.0 \mathrm{~m/s}\). We need to find out how long it takes for Car A to catch up with Car B after Car A starts accelerating.
02

Set Up the Equations

First, we use the equation of motion for Car A: \( s_A = \frac{1}{2} a t^2 \), where \(s_A\) is the distance covered by Car A, \(a = 6.0 \mathrm{~m/s^2}\) is the acceleration, and \(t\) is the time in seconds. For Car B, which moves at constant speed, the equation is \( s_B = v t \), where \(v = 70.0 \mathrm{~m/s}\) is the speed.
03

Set the Distances Equal

Since Car A needs to catch up with Car B, we set their distances equal: \( \frac{1}{2} a t^2 = v t \). Substitute \(a = 6.0 \mathrm{~m/s^2}\) and \(v = 70.0 \mathrm{~m/s}\) into the equation: \( \frac{1}{2} \, \times \, 6.0 \, t^2 = 70.0 \, t \).
04

Solve the Equation for Time

Cancel a \(t\) from both sides (assuming \( t eq 0 \)), which gives: \( 3.0 t = 70.0 \). Now, solve for \(t\): \( t = \frac{70.0}{3.0} = 23.33 \) seconds.
05

Verify the Answer

Calculate the distance each car travels during this time. Plug \(t = 23.33\) into \( s_A = \frac{1}{2} \times 6.0 \times (23.33)^2 \) to find the distance traveled by Car A. Check that it matches the distance traveled by Car B, which should be \( s_B = 70.0 \times 23.33 \). Both should equal therefore, the time calculated is correct.

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Key Concepts

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

acceleration
Acceleration is a measure of how quickly an object speeds up or slows down. It is a vector quantity, which means it has both magnitude and direction. For an object that starts from rest and accelerates constantly, as in the case of Car A in the exercise, its acceleration is given by the rate of change of velocity over time.
The formula to calculate acceleration is:
  • \( a = \frac{\Delta v}{t} \)
where \( a \) is the acceleration, \( \Delta v \) is the change in velocity, and \( t \) is the time over which the change occurs.
In the exercise, Car A has an acceleration of \(6.0 \mathrm{~m/s^2}\), which means each second it gains \(6.0 \mathrm{~m/s}\) in speed as long as it accelerates. Understanding acceleration helps predict how the speed of an object will increase over time.
constant speed
Constant speed means that an object travels the same distance every second. This is the condition for Car B in the exercise. Unlike acceleration, constant speed implies no change in speed over time.
The formula to determine the distance covered at a constant speed is:
  • \( s = v \times t \)
where \( s \) is the distance traveled, \( v \) is the constant speed, and \( t \) is the time of travel.
For Car B, moving at \(70.0 \mathrm{~m/s}\), it covers \(70.0\) meters every second. This steady pace is a great counterpoint to Car A's increasing speed, offering a clear view of how different speed conditions interact in motion.
equations of motion
Equations of motion are mathematical formulas used to describe the motion of an object under certain conditions. They can solve problems involving initial velocity, final velocity, acceleration, time, and displacement. They are essential tools in kinematics—the study of motion.
For uniformly accelerated motion, the main equations are:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2} at^2 \)
  • \( v^2 = u^2 + 2as \)
where \( u \) is the initial velocity, \( v \) is the final velocity, \( a \) is the acceleration, \( s \) is the displacement, and \( t \) is the time.
In the given problem, we use the second equation to determine the distance Car A covers as it accelerates – starting from rest with a given acceleration over a defined time.
relative motion
Relative motion is a concept used to describe the motion of an object as observed from another object. It's crucial in understanding how different frames of reference affect motion.
In the exercise, the motion of Car A is examined relative to Car B. The goal is for Car A to reach the same point as Car B and match its speed, even though they started with different initial conditions.
To solve such problems, it's helpful to consider:
  • The position or distance each object has traveled since they started moving.
  • The time it takes for the relative positions to equal each other.
  • The speeds or accelerations, showing how an object increases or maintains its velocity compared to another object's movement.
Relative motion calculations often simplify complex problems by focusing on the differences in velocity, position, and time between objects. By understanding these concepts, predicting when and where two moving objects will intersect becomes straightforward.

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Most popular questions from this chapter

A speedboat starts from rest and accelerates at \(+2.01 \mathrm{~m} / \mathrm{s}^{2}\) for \(7.00 \mathrm{~s}\). At the end of this time, the boat continues for an additional \(6.00 \mathrm{~s}\) with an acceleration of \(+0.518 \mathrm{~m} / \mathrm{s}^{2}\) Following this, the boat accelerates at \(-1.49 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s}\). (a) What is the velocity of the boat at \(t=21.0 \mathrm{~s} ?\) (b) Find the total displacement of the boat.

Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, motorcycle A has an average acceleration of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) due east, while motorcycle \(\mathrm{B}\) has an average acceleration of \(4.0 \mathrm{~m} / \mathrm{s}^{2}\) due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster?

A whale swims due east for a distance of \(6.9 \mathrm{~km},\) turns around and goes due west for \(1.8 \mathrm{~km},\) and finally turns around again and heads \(3.7 \mathrm{~km}\) due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction of the displacement of the whale?

A ball is thrown upward from the top of a 25.0 -m-tall building. The ball's initial speed is \(12.0 \mathrm{~m} / \mathrm{s}\). At the same instant, a person is running on the ground at a distance of \(31.0 \mathrm{~m}\) from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

A cement block accidentally falls from rest from the ledge of a 53.0 -m-high building. When the block is \(14.0 \mathrm{~m}\) above the ground, a man, \(2.00 \mathrm{~m}\) tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

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