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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, travcling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s},\) overtakes and passcs 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 meters; (b) 9.33 m/s.

Step by step solution

01

Determine the Distance Traveled by the Automobile

The automobile starts from rest and moves with a constant acceleration. The distance traveled by the automobile can be calculated using the equation for motion:\[ s = ut + \frac{1}{2}at^2 \]where, \( u = 0 \) (initial velocity),\( a = 4.0 \ \text{m/s}^2 \) (acceleration),and \( t = 3.0 \ \text{s} \) (time).Substitute the values:\[ s = 0 \times 3 + \frac{1}{2} \times 4.0 \times 3^2 = 18 \ \text{m} \]So, the automobile travels a distance of 18 meters.
02

Determine the Distance Traveled by the Truck

The truck moves at a constant speed, so the distance it travels is given by:\[ s = vt \]where,\( v = 8.0 \ \text{m/s} \) (velocity) and\( t = 3.0 \ \text{s} \) (time).Substitute the values:\[ s = 8.0 \times 3 = 24 \ \text{m} \]Thus, the truck travels a distance of 24 meters.
03

Calculate the Center of Mass of the System

The position of the center of mass \( x_{com} \) is given by:\[ x_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]where,- \( m_1 = 1000 \ \text{kg} \) and \( x_1 = 18 \ \text{m} \) (position of the automobile),- \( m_2 = 2000 \ \text{kg} \) and \( x_2 = 24 \ \text{m} \) (position of the truck).Substitute the values:\[ x_{com} = \frac{1000 \times 18 + 2000 \times 24}{1000 + 2000} = \frac{18000 + 48000}{3000} = 22 \ \text{m} \]Hence, at \( t = 3 \ \text{s} \), the center of mass is 22 meters from the traffic light.
04

Calculate the Speed of the Center of Mass

The speed of the center of mass \( v_{com} \) can be determined using:\[ v_{com} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]where,- \( m_1 = 1000 \ \text{kg} \) and \( v_1 = a \cdot t = 4.0 \ \text{m/s}^2 \times 3.0 \ \text{s} = 12 \ \text{m/s} \) (velocity of the automobile at \( t = 3 \ \text{s} \)),- \( m_2 = 2000 \ \text{kg} \) and \( v_2 = 8.0 \ \text{m/s} \) (velocity of the truck).Substitute the values:\[ v_{com} = \frac{1000 \times 12 + 2000 \times 8}{3000} = \frac{12000 + 16000}{3000} = 9.33 \ \text{m/s} \]Therefore, at \( t = 3 \ \text{s} \), the speed of the center of mass of the automobile-truck system is 9.33 m/s.

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Key Concepts

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

Constant Acceleration
When a vehicle like the automobile in the exercise starts from rest and speeds up at a steady pace, it is undergoing constant acceleration. In simple terms, constant acceleration means the vehicle's speed increases by the same amount every second.

Here's how it works in practice: If an automobile accelerates at a rate of \(4.0 \, \text{m/s}^2\), every second its speed increases by \(4.0 \, \text{m/s}\). After one second, its speed is \(4.0 \, \text{m/s}\); after two seconds, it will be \(8.0 \, \text{m/s}\), and so forth. This can be visualized with the equation for constant acceleration:
  • Initial velocity \(u = 0\) because the car starts from rest.
  • Acceleration \(a = 4.0 \, \text{m/s}^2\).
  • Time \(t = 3.0 \, \text{s}\).
  • The formula \(s = ut + \frac{1}{2}at^2\) helps calculate the distance traveled, which in this case is \(18 \, \text{m}\).
This concept is vital in kinematics to predict how objects move over time when starting from rest and experiencing uniform acceleration.
Center of Mass
The center of mass can be likened to the 'balance point' of a system. In the example of the automobile and the truck, the center of mass helps us understand how these two moving objects combine into one system and how we observe their motion collectively.

To find the center of mass of the automobile-truck system, we use the formula:\[x_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]Here, \(m_1\) and \(x_1\) are the mass and distance of the automobile from the starting point, while \(m_2\) and \(x_2\) are for the truck. This allows you to calculate a kind of "weighted average" distance from the traffic light, resulting in \(22 \, \text{m}\) at \(t=3.0 \, \text{s}\).

Understanding the center of mass is essential in classical mechanics since it simplifies the study of motion by reducing a complex system to a simpler model.
Kinematics
Kinematics is the branch of classical mechanics that describes the motion of objects without necessarily considering the forces causing the motion. It's all about understanding three key aspects: displacement, velocity, and acceleration.

Using kinematics, we can predict how far the automobile and truck travel and at what speed, based simply on their initial conditions and how they move over time:
  • Displacement tells us how far an object has moved; the automobile traveled \(18 \, \text{m}\) and the truck \(24 \, \text{m}\).
  • Velocity refers to the speed of the objects involved; at \(t = 3.0 \, \text{s}\), the automobile moves at \(12 \, \text{m/s}\) and the truck at a constant \(8.0 \, \text{m/s}\).
  • Acceleration specifies the change in velocity; in this case, only the automobile accelerates at \(4.0 \, \text{m/s}^2\).
By combining these elements, kinematics aids in developing a clear understanding of the motion of the automobile and truck from the moment the light turns green.

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Most popular questions from this chapter

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