Problem 26 - Calculus Single Variable

Question

A car accelerates from rest at the constant rate of \(2 \mathrm{mi} / \mathrm{min}^{2}\). After the car has traveled one eighth of a mile, how fast will it be traveling?

Step-by-Step Solution

Answer

The car will be traveling approximately 0.707 mi/min.

Step 1: Understanding the Problem

We're given a car that starts from rest and accelerates at a constant rate of \(2 \text{ mi/min}^2\). We need to determine the car's velocity after it has traveled \(\frac{1}{8}\) mile.

Step 2: Setting up the Known Values

Initial velocity \(u = 0 \text{ mi/min}\), acceleration \(a = 2 \text{ mi/min}^2\), and the distance traveled \(s = \frac{1}{8} \text{ mi}\).

Step 3: Applying the Equation of Motion

Use the second equation of motion: \(v^2 = u^2 + 2as\), where \(v\) is the final velocity. Substitute the known values: \(u = 0\), \(a = 2\text{ mi/min}^2\), \(s = \frac{1}{8} \text{ mi}\).

Step 4: Substituting the Values

Substitute into the formula: \(v^2 = 0 + 2 \times 2 \times \frac{1}{8}\). Simplify to find \(v^2 = \frac{4}{8} = \frac{1}{2}\).

Step 5: Solving for the Final Velocity

Take the square root of both sides to find \(v: v = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707 \text{ mi/min}\).

Key Concepts

Understanding Acceleration

Acceleration is the rate of change of velocity over time. In a motion problem, when an object changes its speed, it is accelerating. The unit of acceleration in this problem is miles per minute squared (mi/min²), showing how quickly the speed of the car increases each minute. For instance, a constant acceleration of 2 mi/min² means that every minute, the car speeds up by 2 miles per minute.

• Acceleration is a vector, meaning it has both magnitude and direction.
• Constant acceleration implies the rate of change of speed is unvarying.
• Understanding acceleration is key to solving motion problems, as it determines how quickly an object can change its speed.

To calculate the effect of acceleration over a distance, we use specific equations of motion, which we'll explore next.

Exploring Equations of Motion

Equations of motion are formulas that relate the various physical quantities involved in moving objects — like initial velocity, final velocity, acceleration, time, and distance. In this exercise, we used the second equation of motion: \[v^2 = u^2 + 2as\]This equation is very useful when time is not directly involved in the calculation. It allows us to find the final velocity based on the initial velocity, acceleration, and distance traveled. Let's break down each component:

• • v is the final velocity we want to find.
  • u is the initial velocity — in our scenario, it's 0 mi/min as the car starts from rest.
  • a is the acceleration — we know it's 2 mi/min².
  • s is the distance traveled — given as 1/8 mile in this problem.

By substituting these known values into the equation, we can easily compute the car's final velocity.

Grasping Velocity

Velocity is a measure of how fast something is moving, in a specific direction. It's different from speed because velocity is a vector quantity — it considers direction as well as magnitude. In this motion problem, we calculated the final velocity to understand how quickly the car moves after traveling one-eighth of a mile.

• Velocity at rest is zero; here the car begins from rest so the initial velocity u is 0 mi/min.
• The final velocity (v) determined from calculations tells us how fast the car is moving after covering a certain distance with a given acceleration.
• The calculated velocity, about 0.707 mi/min, gives a precise measure of the car's movement at that point in its acceleration.

Velocity is crucial for understanding motion as it gives us insight into the dynamics of moving objects, particularly in determining how fast they're getting from one point to another.