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Chapter 2: Problem 62

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of \(2.50 \mathrm{~m} / \mathrm{s}^{2}\). At the same instant, a truck, traveling with a constant speed of \(15.0 \mathrm{~m} / \mathrm{s}\), overtakes and passes the automobile. (a) How far beyond its starting point does the automobile overtake the truck? (b) How fast is the automobile traveling when it overtakes the truck?

Short Answer

Expert verified
(a) 180 m, (b) 30 m/s.

Step by step solution

01

Define the Problem

We need to find when the distance traveled by the accelerating car equals the distance traveled by the truck. Let the distance be \( x \). The acceleration of the automobile is \( a = 2.50 \, \mathrm{m/s}^{2} \), and the speed of the truck is \( v = 15.0 \, \mathrm{m/s} \).
02

Write Equations for Distance

The distance traveled by the automobile can be calculated using the equation \( x = \frac{1}{2}at^2 \), where \( a = 2.50 \, \mathrm{m/s}^2 \). The distance traveled by the truck is \( x = vt \), where \( v = 15.0 \, \mathrm{m/s} \).
03

Equate the Distances

Set the distance equations equal to each other: \( \frac{1}{2}at^2 = vt \). Substitute \( a = 2.50 \, \mathrm{m/s}^2 \) and \( v = 15.0 \, \mathrm{m/s} \) into the equation: \( \frac{1}{2}(2.50)t^2 = 15t \).
04

Solve for Time (t)

Rearrange the equation \( 1.25t^2 = 15t \) to \( 1.25t^2 - 15t = 0 \). Factor the equation: \( t(1.25t - 15) = 0 \). The solutions are \( t = 0 \) or \( 1.25t = 15 \). Solve \( 1.25t = 15 \) to find \( t = 12 \; \mathrm{s} \).
05

Calculate Distance

Use the time from Step 4 in the equation \( x = vt \) or \( x = \frac{1}{2}at^2 \). Using \( x = vt \), we get \( x = 15 \times 12 = 180 \, \mathrm{m} \).
06

Calculate Automobile's Speed at Overtake

Use the speed formula for uniformly accelerated motion: \( v = u + at \). Here, \( u = 0 \) and \( a = 2.50 \, \mathrm{m/s}^2 \), so \( v = 0 + 2.5 \times 12 = 30 \mathrm{~m/s} \).

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Key Concepts

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

Constant acceleration
In physics, **constant acceleration** refers to a situation where an object's velocity changes by the same amount in every equal time period. This is an important concept because it allows us to predict how objects move over time when they are subjected to constant forces. For example, a car accelerating from a stop with a constant acceleration of \(2.50 \, \mathrm{m/s}^2\) means that every second, its speed increases by \(2.50 \, \mathrm{m/s}\).
This type of motion is described using equations from kinematics, such as:
  • \(v = u + at\): Final velocity (\(v\)) is the initial velocity (\(u\)) plus the product of acceleration (\(a\)) and time (\(t\)).
  • \(s = ut + \frac{1}{2}at^2\): The distance (\(s\)) travelled under constant acceleration can be found using this formula where \(u\) is the initial velocity, \(a\) is acceleration, and \(t\) is time.
Both equations can help us solve various problems involving constant acceleration, such as determining speed, distance, or time.
Uniform motion
**Uniform motion** refers to motion at a constant speed in a straight line. This means there is no acceleration involved, and an object covers equal distances in equal intervals of time. For instance, a truck traveling at \(15.0 \, \mathrm{m/s}\) exhibits uniform motion since its speed does not change over time.
To calculate the distance travelled under uniform motion, we use the simple equation:
  • \(s = vt\): Here, \(s\) is the distance, \(v\) is the constant velocity, and \(t\) is the time.
Understanding uniform motion is crucial for solving problems where an object's speed remains the same over time, as seen when the truck in our problem maintains a constant speed throughout its journey.
Equating distances
In physics word problems, **equating distances** is a common method used to find the point at which two moving objects meet or overtake each other. This involves setting the distance travelled by the first object equal to the distance travelled by the second object.
In the given exercise, we set the distance the car travels while accelerating equal to the distance the truck covers moving at a constant speed. The equations used are:
  • For the car: \(x = \frac{1}{2}at^2\)
  • For the truck: \(x = vt\)
By equating these two expressions, we can solve for the time \(t\) at which both have covered the same distance. This method leverages their respective equations of motion to find out when and where their paths converge.
Solving quadratic equations
**Solving quadratic equations** is an essential skill in solving physics problems, especially when the laws of motion are involved. Quadratic equations are polynomial equations of degree 2, generally in the form of \(ax^2 + bx + c = 0\).
In our scenario, setting the equations for the distance travelled by both the car and the truck leads us to a quadratic equation \(1.25t^2 - 15t = 0\). This can be solved using:
  • Factoring: We rearranged to \(t(1.25t - 15) = 0\). This gives possible solutions \(t = 0\) or \(t = 12\) seconds, choosing the non-zero solution.
  • The Quadratic Formula: For equations that aren't easily factored, use \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Solving these equations allows us to find when the automobile overtakes the truck, which happens at \(t = 12\) seconds in this problem.

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Most popular questions from this chapter

On March \(27,2004,\) the United States successfully tested the hypersonic X-43A scramjet, which flew at Mach 7 (seven times the speed of sound) for 11 s. (a) At this rate, how many minutes would it take such a scramjet to carry passengers the approximately \(5000 \mathrm{~km}\) from San Francisco to New York? (Use the speed of sound at \(\left.0^{\circ} \mathrm{C}, 331 \mathrm{~m} / \mathrm{s} .\right)\) (b) How many kilometers did the scramjet travel during its 11 s test?

If the contraction of the left ventricle lasts \(250 \mathrm{~ms}\) and the speed of blood flow in the aorta (the large artery leaving the heart) is \(0.80 \mathrm{~m} / \mathrm{s}\) at the end of the contraction, what is the average acceleration of a red blood cell as it leaves the heart? A. \(310 \mathrm{~m} / \mathrm{s}^{2}\) B. \(31 \mathrm{~m} / \mathrm{s}^{2}\) C. \(3.2 \mathrm{~m} / \mathrm{s}^{2}\) D. \(0.32 \mathrm{~m} / \mathrm{s}^{2}\)

Two runners start simultaneously at opposite ends of a \(200.0 \mathrm{~m}\) track and run toward each other. Runner \(A\) runs at a steady \(8.0 \mathrm{~m} / \mathrm{s}\) and runner \(B\) runs at a constant \(7.0 \mathrm{~m} / \mathrm{s}\). When and where will these runners meet?

Suppose you are climbing in the High Sierra when you suddenly find yourself at the edge of a fogshrouded cliff. To find the height of this cliff, you drop a rock from the top and, \(10.0 \mathrm{~s}\) later, hear the sound of it hitting the ground at the foot of the cliff. (a) Ignoring air resistance, how high is the cliff if the speed of sound is \(330 \mathrm{~m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.

Cheetahs, the fastest of the great cats, can reach \(45 \mathrm{mi} / \mathrm{h}\) in \(2.0 \mathrm{~s}\) starting from rest. Assuming that they have constant acceleration throughout that time, find (a) their acceleration (in \(\mathrm{ft} / \mathrm{s}^{2}\) and \(\left.\mathrm{m} / \mathrm{s}^{2}\right)\) and \((\mathrm{b})\) the distance (in \(\mathrm{m}\) and \(\mathrm{ft}\) ) they travel during that time.
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