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A \(1000 \mathrm{~kg}\) automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of \(4.0 \mathrm{~m} / \mathrm{s}^{2}\). At the same instant a \(2000 \mathrm{~kg}\) truck, traveling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\), overtakes and passes the automobile. (a) How far is the com of the automobile-truck system from the traffic light at \(t=3.0 \mathrm{~s}\) ? (b) What is the speed of the com then?

Short Answer

Expert verified
(a) 22.0 m, (b) 9.33 m/s

Step by step solution

01

Understand System Components

The problem involves two vehicles: a car with mass 1000 kg starting from rest with constant acceleration, and a truck with mass 2000 kg maintaining constant speed. Our task is to find the position and speed of the center of mass (COM) after 3 seconds.
02

Determine Motion Equations

For the car, we use the equation of motion for uniformly accelerated motion:\[ x_{car} = ut + \frac{1}{2}at^2 \]where \( u = 0 \) (initial velocity), \( a = 4.0 \, \mathrm{m/s}^2 \), and \( t = 3 \) seconds. For the truck, moving at constant speed:\[ x_{truck} = v_{truck} \times t \]where \( v_{truck} = 8.0 \mathrm{~m/s} \) and \( t = 3 \) seconds.
03

Calculate Position of Each Vehicle

Substituting into the equations:For the car:\[ x_{car} = 0 + \frac{1}{2} \times 4.0 \times (3)^2 = 18.0 \mathrm{~m} \]For the truck:\[ x_{truck} = 8.0 \times 3 = 24.0 \mathrm{~m} \]
04

Find Position of COM

The position of the center of mass \( x_{COM} \) is given by:\[ x_{COM} = \frac{m_{car} \times x_{car} + m_{truck} \times x_{truck}}{m_{car} + m_{truck}} \]Substituting the known values:\[ x_{COM} = \frac{1000 \times 18.0 + 2000 \times 24.0}{1000 + 2000} = \frac{18000 + 48000}{3000} = 22.0 \mathrm{~m} \]
05

Calculate Speed of Each Vehicle

For the car, the speed at \( t = 3 \) seconds is found using:\[ v_{car} = u + at = 0 + 4.0 \times 3 = 12.0 \mathrm{~m/s} \]The truck’s speed \( v_{truck} = 8.0 \mathrm{~m/s} \) (constant).
06

Determine Speed of COM

The velocity of the center of mass \( v_{COM} \) is given by:\[ v_{COM} = \frac{m_{car} \times v_{car} + m_{truck} \times v_{truck}}{m_{car} + m_{truck}} \]Substituting the known values:\[ v_{COM} = \frac{1000 \times 12.0 + 2000 \times 8.0}{1000 + 2000} = \frac{12000 + 16000}{3000} = 9.33 \mathrm{~m/s} \]

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Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion form the bedrock for understanding dynamics, motion, and how objects interact with forces in our universe. The first law, often referred to as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This clearly applies to the truck in our exercise, which continues at a constant speed without additional forces acting upon it.

Newton's Second Law of Motion explains how the velocity of an object changes when it is subjected to an external force. It introduces the equation: \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. For the car in our exercise, an acceleration of \( 4.0 \, ext{m/s}^2 \) is applied, demonstrating this law in action as it moves from rest.

The third law, stating that every action has an equal and opposite reaction, describes how the car's wheels push against the road propelling it forward, while the road pushes back with an equal force.
Acceleration
Acceleration is a crucial concept in understanding how objects change their velocity over time. When an object accelerates, it means its speed is increasing, decreasing, or its direction is changing. In our scenario, the car illustrates uniform acceleration, starting from rest and continuously increasing its speed due to a steady force applied over time. This kind of motion is described by the formula: \( x = ut + \frac{1}{2}at^2 \) where \( x \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is time.

It's important to note that even though the car's speed is constantly increasing, the acceleration remains constant at \( 4.0 \, ext{m/s}^2 \). This means every second, the car's velocity increases by an additional 4 meters per second, a perfect example of uniform acceleration in physics.

Acceleration can significantly affect the motion of an object's center of mass, which in turn impacts the composite speed and trajectory of systems involving multiple bodies, like the car-truck system in our exercise.
Uniform Motion
Uniform motion refers to the movement of an object at a constant speed in a straight line. It implies zero acceleration, meaning the object's velocity is unchanging. This is what the truck in our exercise demonstrates. Regardless of other forces around it, the truck's speed is unaffected and constantly measured at \( 8.0 \, ext{m/s} \).

Understanding uniform motion is essential when calculating the motion of multiple objects, as it simplifies the determination of distances traveled and helps predict future positions. The equation for uniform motion is simply \( x = vt \), where \( x \) is the distance traveled, \( v \) is the constant velocity, and \( t \) is the time elapsed.

In the case of our vehicle problem, the truck overtakes the car immediately because it maintains its initial speed, illustrating how uniform motion leads to straightforward calculations in physics, contrasting with the car's uniformly accelerating movement which involves more complex computational steps.

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Most popular questions from this chapter

Figure 9-51 shows a 0.300 \(\mathrm{kg}\) baseball just before and just after it collides with a bat. Just before, the ball has velocity \(\vec{v}_{1}\) of magnitude \(12.0 \mathrm{~m} / \mathrm{s}\) and angle \(\theta_{1}=35.0^{\circ}\). Just after, it is traveling directly upward with velocity \(\vec{v}_{2}\) of magnitude \(10.0\) \(\mathrm{m} / \mathrm{s}\). The duration of the collision is \(2.00 \mathrm{~ms}\). What are the (a) magnitude and (b) direction (relative to the positive direction of the \(x\) axis) of the impulse on the ball from the bat? What are the (c) magnitude and (d) direction of the average force on the ball from the bat?

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A \(0.25 \mathrm{~kg}\) puck is initially stationary on an ice surface with negligible friction. At time \(t=0\), a horizontal force begins to move the puck. The force is given by \(\vec{F}=\left(12.0-3.00 t^{2}\right) \hat{1}\), with \(\vec{F}\) in newtons and \(t\) in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between \(t=0.500 \mathrm{~s}\) and \(t=1.25 \mathrm{~s} ?\) (b) What is the change in momentum of the puck between \(t=0\) and the instant at which \(F=0 ?\)

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