/*! 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 66 At the instant the traffic light... [FREE SOLUTION] | 91Ó°ÊÓ

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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) The automobile overtakes the truck 180 meters beyond its starting point. (b) The car's speed at that moment is 30.0 m/s.

Step by step solution

01

Define Variables

Let's define the variables first: - Acceleration of the car, \( a = 2.50 \text{ m/s}^2 \).- Velocity of the truck, \( v_t = 15.0 \text{ m/s} \).- Initial velocity of the car is \( 0 \text{ m/s} \).We need to find the distance \( x \) from the starting point where the car overtakes the truck, and the speed \( v_c \) of the car at that moment.
02

Set Equations for Distances

The distance covered by the car \( x_c \) is given by:\[x_c = \frac{1}{2} a t^2\]The distance covered by the truck \( x_t \) is given by:\[x_t = v_t \times t\]The car overtakes the truck when these distances are equal, so we set:\[\frac{1}{2} a t^2 = v_t \times t\]
03

Solve for Time \(t\)

Rearrange the equation to solve for \( t \):\[\frac{1}{2} a t = v_t\]\[t = \frac{2v_t}{a}\]Substitute known values:\[t = \frac{2 \times 15.0}{2.50} = 12 \text{ s}\]
04

Calculate Distance \(x\)

Substitute \( t = 12 \) seconds back into the equation for the car’s distance:\[x_c = \frac{1}{2} \times 2.50 \times (12)^2\]\[x_c = \frac{1}{2} \times 2.50 \times 144 = 180 \text{ m}\]So, the car overtakes the truck at 180 meters from the starting point.
05

Determine Car's Speed \(v_c\) When Overtaking

The speed of the car \( v_c \) when it overtakes the truck can be found using:\[v_c = u + at\]Since the initial speed \( u = 0 \), we have:\[v_c = 0 + 2.50 \times 12\]\[v_c = 30.0 \text{ m/s}\]Thus, the car is traveling at a speed of 30.0 m/s when it overtakes the truck.

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Key Concepts

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

Constant Acceleration
In the realm of physics, constant acceleration occurs when an object's velocity changes at a steady rate over time. This means that the object's speed increases or decreases uniformly with every passing second. For instance, in our exercise, the car starts accelerating from rest with a constant acceleration of 2.50 m/s².
To understand how constant acceleration affects motion, consider an object under these conditions. Its velocity at any time can be calculated using the formula:
  • The velocity after time \( t \) is given by \( v = u + at \), where \( u \) is the initial velocity and \( a \) is the acceleration.
This concept is crucial in physics problems involving speeds and movements, like determining how fast an automobile is traveling or when it overtakes another vehicle. By applying these formulas, we simplify complex real-world problems into manageable calculations.
Physics Problems
Physics problems often involve applying mathematical formulas to understand and predict the real-world movements of objects. By solving exercises like the one given, students practice translating problems into mathematical equations.
Initially, define the variables involved for clarity. For instance, identify the initial velocities, acceleration rates, and time required for changes to occur. With these, write down the relevant equations that describe the system's dynamics:
  • For the car: \( x_c = \frac{1}{2} a t^2 \)
  • For the truck: \( x_t = v_t \times t \)
These equations reflect the distance covered by each vehicle, allowing you to solve for unknown values such as time \( t \) or distance \( x \).
By understanding the setup and relationships between variables, you can tackle more complex problems with confidence.
Kinematics
Kinematics is the study of motion without considering the forces that cause it. It focuses on specific parameters like velocity, acceleration, and time. In our exercise, kinematics helps explain how and when the car overtakes the truck.
Using kinematics equations, you can determine how objects move based on given parameters. For example, to find out when an accelerating car overtakes a truck moving at constant speed, equate the distances traveled by both:
  • The car's distance is expressed as \( x_c = \frac{1}{2} a t^2 \)
  • The truck’s constant speed means its distance is \( x_t = v_t \times t \)
By setting these equal, the time \( t \) when the car overtakes the truck is derived. Kinematics provides the tools to draw accurate conclusions about motion scenarios.
Velocity
Velocity is a measure of how fast an object is moving in a specific direction. Unlike speed, it is a vector quantity, meaning it includes both a magnitude (how fast) and a direction (where to).
In the given problem, understanding velocity informs us of the car's behavior as it catches up to the truck. Initially, the car is stationary, meaning its starting velocity is 0 m/s. As it accelerates, its velocity increases.
  • To determine the car's velocity when it overtakes the truck, use the formula: \( v_c = u + at \)
  • Given that \( u = 0 \), this simplifies to \( v_c = at \)
  • At \( t = 12 \) seconds, the velocity is calculated as \( v_c = 2.50 \times 12 = 30.0 \text{ m/s} \)
Understanding changes in velocity helps predict an object's future position and timing during motion.

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