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A top-fuel dragster starts from rest and has a constant acceleration of \(40.0 \mathrm{m} / \mathrm{s}^{2}\). Concepts: (i) At time \(t\) the dragster has a certain velocity. Keeping in mind that the dragster starts from rest, when the time doubles to \(2 t\), does the velocity also double? (ii) When the time doubles to \(2 t,\) does the displacement of the dragster also double? Calculations: What are the (a) final velocities and (b) displacements of the dragster at the end of \(2.0 \mathrm{s}\) and at the end of twice this time, or \(4.0 \mathrm{s} ?\)

Short Answer

Expert verified
(i) Yes, velocity doubles. (ii) No, displacement quadruples.

Step by step solution

01

Identify Given Information and Formulas

We know the dragster starts from rest, so initial velocity \( u = 0 \). The acceleration \( a = 40.0 \ \text{m/s}^2 \). We need to find final velocities and displacements for times \( t = 2.0 \ \text{s} \) and \( 2t = 4.0 \ \text{s} \). Relevant formulas are the equations of motion: \( v = u + at \) for velocity and \( s = ut + \frac{1}{2}at^2 \) for displacement.
02

Calculate Velocity at t = 2.0 s

Using the formula \( v = u + at \), substitute \( u = 0 \), \( a = 40.0 \ \text{m/s}^2 \), and \( t = 2.0 \ \text{s} \): \[ v = 0 + (40.0)(2.0) = 80.0 \ \text{m/s} \].The final velocity at \( t = 2.0 \ \text{s} \) is \( 80.0 \ \text{m/s} \).
03

Calculate Displacement at t = 2.0 s

Using the formula \( s = ut + \frac{1}{2}at^2 \), plug in \( u = 0 \), \( a = 40.0 \ \text{m/s}^2 \), and \( t = 2.0 \ \text{s} \):\[ s = (0)(2.0) + \frac{1}{2}(40.0)(2.0)^2 = 80.0 \ \text{m} \].The displacement at \( t = 2.0 \ \text{s} \) is \( 80.0 \ \text{m} \).
04

Calculate Velocity at t = 4.0 s

Using the same velocity formula, substitute \( t = 4.0 \ \text{s} \): \[ v = 0 + (40.0)(4.0) = 160.0 \ \text{m/s} \].The final velocity at \( t = 4.0 \ \text{s} \) is \( 160.0 \ \text{m/s} \).
05

Calculate Displacement at t = 4.0 s

For displacement at \( t = 4.0 \ \text{s} \), use \( s = ut + \frac{1}{2}at^2 \):\[ s = (0)(4.0) + \frac{1}{2}(40.0)(4.0)^2 = 320.0 \ \text{m} \].The displacement at \( t = 4.0 \ \text{s} \) is \( 320.0 \ \text{m} \).
06

Analyze the Doubling Time Effect

For question (i), the velocity at \( t = 4.0 \ \text{s} \) is double the velocity at \( t = 2.0 \ \text{s} \), confirming that velocity doubles when time doubles.For question (ii), the displacement at \( t = 4.0 \ \text{s} \) is four times the displacement at \( t = 2.0 \ \text{s} \), indicating displacement does not simply double with time.

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Key Concepts

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

Velocity and Time
In physics, understanding the relationship between velocity and time is fundamental, especially when considering motion with constant acceleration. When an object starts from rest, its initial velocity is zero. For a dragster starting from rest with a constant acceleration of 40.0 m/s², its velocity will increase steadily over time. The formula that helps us calculate this velocity is: \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity (zero in this case), \( a \) is the acceleration, and \( t \) is the time.

For instance, at \( t = 2 \) seconds, the dragster's velocity is \( 80.0 \) m/s. When the time doubles to \( t = 4 \) seconds, the velocity becomes \( 160.0 \) m/s. This clearly illustrates that when time doubles, the velocity doubles as well, provided that the acceleration remains constant.

This constant rate of change implies a linear relationship between velocity and time, a crucial insight for mastering the equations of motion.
Displacement and Time
Displacement refers to how far an object has traveled from its starting point. It's important to realize that the relationship between displacement and time isn't as straightforward as with velocity. The formula for displacement under constant acceleration is: \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

In our dragster example, at \( t = 2 \) seconds, the displacement is \( 80.0 \) meters. However, when time is doubled to \( t = 4 \) seconds, the displacement becomes \( 320.0 \) meters. This shows that displacement quadruples even though time only doubles.

This quadratic relationship is due to the \( t^2 \) term in the displacement formula, indicating that displacement grows much more rapidly than the linear change of velocity. Understanding this concept is key to solving problems involving moving objects.
Constant Acceleration
Constant acceleration means that the rate of change of velocity remains the same over time. It's a central concept when using the equations of motion to describe the behavior of moving objects. In scenarios like our dragster, the constant acceleration is \( 40.0 \) m/s², meaning the velocity increases by \( 40.0 \) m/s every second.

- Key Equations:
  • Velocity: \( v = u + at \)
  • Displacement: \( s = ut + \frac{1}{2}at^2 \)
Constant acceleration allows us to make predictions about an object's future velocity and position, which is essential for applications in engineering, physics, and everyday situations.

Remember, while constant acceleration leads to linear changes in velocity, it results in quadratic changes in displacement. This makes understanding this concept crucial in physics and real-world motion analysis.

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Most popular questions from this chapter

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