Problem 27 - Calculus Single Variable

Question

A sprinter accelerates during the first \(20 \mathrm{~m}\) of a race at the rate of \(4 \mathrm{~m} / \mathrm{s}^{2}\). Of course, she begins at rest. How fast is she running at the moment she hits the \(20 \mathrm{~m}\) mark?

Step-by-Step Solution

Answer

The sprinter is running at approximately 12.65 m/s when she reaches the 20-meter mark.

Step 1: Understanding Given Data

We know that the sprinter begins at rest, which means the initial velocity \(u\) is zero. The sprinter accelerates over a distance \(s = 20 \text{ m}\) with an acceleration \(a = 4 \text{ m/s}^2\). We need to find the final velocity \(v\) at the 20-meter mark.

Step 2: Recognizing the Equation to Use

To find the final velocity, we use the kinematic equation: \(v^2 = u^2 + 2as\). This formula relates the initial velocity, final velocity, acceleration, and distance.

Step 3: Substituting Known Values

Since \(u = 0\), the equation simplifies to \(v^2 = 2as\). Substitute the given values: \(v^2 = 2 \times 4 \times 20\).

Step 4: Calculating Final Velocity

Calculate \(v^2\): \(v^2 = 160\). Find \(v\) by taking the square root of both sides: \(v = \sqrt{160}\).

Step 5: Simplifying the Final Calculation

Compute \(\sqrt{160} = 4\sqrt{10}\), approximately \(12.65 \text{ m/s}\). The sprinter is running at 12.65 meters per second when she hits the 20-meter mark.

Key Concepts

Understanding Acceleration

Acceleration is a key concept in kinematics, which deals with motion-related quantities. Acceleration refers to the rate of change of velocity of an object with respect to time.
It essentially tells us how quickly something is speeding up or slowing down. If a sprinter accelerates at 4 meters per second squared, it means that every second, their velocity increases by 4 meters per second. When an object accelerates, different forces might be at play:

• Positive acceleration: Speeding up in a specified direction.
• Negative acceleration (deceleration): Slowing down.

Calculating acceleration involves knowing the change in velocity and the time it takes for that change. The general formula is given by: \[ a = \frac{\Delta v}{\Delta t} \] where \( a \) is the acceleration, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the time period over which the change occurs. Understanding this concept will help you analyze motion problems, and in many cases like the sprinter exercise, you will have a constant acceleration due to the applied force.

Exploring Velocity

Velocity is a vector quantity that refers to the rate at which an object changes its position. Unlike speed, which is scalar, velocity also takes direction into account. This means that knowing an object's speed is not enough; we need to know its direction too.
There are a couple of important points to remember about velocity:

• Initial Velocity \( u \): The velocity at which motion begins. In the sprinter example, the initial velocity is 0, as she starts from rest.
• Final Velocity \( v \): The velocity at the end of motion or after a certain period of time or distance. This is often the quantity we are solving for using kinematic equations.

The velocity can change due to acceleration or changes in direction. For a constant acceleration, velocity can be calculated using the formula: \[ v = u + at \] where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. For distance-based calculations and when time is not directly involved, we use the kinematic equation which connects initial velocity, final velocity, acceleration, and distance, as seen in the sprinter problem.

Unraveling Kinematic Equations

Kinematic equations are foundational tools in motion analysis, especially when forces act in a consistent and predictable manner, like constant acceleration. These equations help us to solve for unknown variables like velocity, distance, acceleration, or time when some variables are known. The four key kinematic equations include:

• \( v = u + at \)
• \( s = ut + \frac{1}{2}at^2 \)
• \( v^2 = u^2 + 2as \)
• \( s = \frac{1}{2}(u + v)t \)

Each equation has its own specific use scenario. For example, in the sprinter problem, we care about motion over a distance with a known acceleration and no given time, so we use: \[ v^2 = u^2 + 2as \] Since the initial velocity \( u \) is zero in the beginning, the equation simplifies, making it easier to calculate the final velocity without time information. By substituting the known values directly into the formula, you can find the unknown, demystifying even complex motion scenarios. These equations are essential in physics as they provide a powerful means to find relationships between various motion aspects.