Problem 12 - Basic Technical Mathematics with Calculus

Question

A person skis down a slope with an acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) given by \(a=\frac{600 t}{\left(60+0.5 t^{2}\right)^{2}},\) where \(t\) is the time (in s). Find the skier's velocity as a function of time if \(v=0\) when \(t=0\)

Step-by-Step Solution

Answer

The skier's velocity as a function of time is \( v(t) = 10 - \frac{600}{60 + 0.5t^2} \).

Step 1: Understand the Given Problem

We need to find the skier's velocity as a function of time. The acceleration is given by \( a = \frac{600t}{(60 + 0.5t^2)^2} \), and the initial velocity \( v = 0 \) when \( t = 0 \).

Step 2: Setup the Integral for Velocity from Acceleration

The relationship between velocity \( v \) and acceleration \( a \) is given by the integral \( v(t) = \int a(t) \, dt \). Substitute the given acceleration function into this integral: \[ v(t) = \int \frac{600t}{(60 + 0.5t^2)^2} \, dt \].

Step 3: Simplify the Integral

Notice that the denominator is a function of \( t^2 \). Set \( u = 60 + 0.5t^2 \), making \( \frac{du}{dt} = t \). This gives us \( du = t \, dt \). The integral becomes: \[ v(t) = 600 \int \frac{1}{u^2} \, du \].

Step 4: Integrate with Respect to u

The integral of \( \frac{1}{u^2} \) is \( -\frac{1}{u} \). So, \[ v(t) = 600 \left(-\frac{1}{u}\right) + C \].

Step 5: Substitute Back for u and Apply Initial Conditions

Substitute back \( u = 60 + 0.5t^2 \) into the equation: \[ v(t) = -\frac{600}{60 + 0.5t^2} + C \]. Using the initial condition \( v(0) = 0 \), obtain \[ 0 = -\frac{600}{60} + C \], which simplifies to \( C = 10 \).

Step 6: Write the Final Velocity Function

Substitute \( C = 10 \) back into the velocity equation: \[ v(t) = -\frac{600}{60 + 0.5t^2} + 10 \]. Simplify to get the final velocity function: \[ v(t) = 10 - \frac{600}{60 + 0.5t^2} \].

Key Concepts

Acceleration

Acceleration describes how quickly an object's velocity changes. It is fundamentally the derivative of velocity with respect to time. In this skiing problem, acceleration is expressed as a function of time:

• Formula: \( a = \frac{600t}{(60 + 0.5t^2)^2} \)
• Units: \( \ ext{m/s}^2 \)

Here, \( a \) changes its value over time based on the function. The skier's acceleration isn’t constant; it depends intricately on the time \( t \), which affects both the speed and the direction of skiing. Understanding how to differentiate is essential in finding such a function, as it details the relation between time and how quickly the speed changes. The more complex an acceleration function, the more nuanced the changes in velocity.

Velocity

Velocity defines the speed and direction of an object. Unlike speed, which is scalar, velocity is a vector quantity. It signifies motion and is derived by integrating acceleration over time. For the skier:

• Initial velocity: The skier starts with a velocity of 0 at time \( t = 0 \).
• Expression as a function of time: \( v(t) = 10 - \frac{600}{60 + 0.5t^2} \).

This solution indicates that the beginning of the motion starts from a standstill (0 velocity). As time progresses, the velocity is adjusted by the determined formula. Velocity being an integral of acceleration highlights the accumulation effect: how past acceleration affects current speed.

Integral Calculus

Integral calculus helps in finding functions when their derivatives are known, which is crucial for problems like these.

• It links acceleration to velocity, allowing us to calculate past changes gathered over time.
• In this exercise, you integrate \( a(t) \) by setting up \( v(t) = \int a(t) \, dt \).
• Substitution Technique: Set \( u = 60 + 0.5t^2 \), simplifying complex integrals.

Integrating acceleration gives you the velocity function. In calculus terms, the integral "adds up" acceleration over time, meaning the formula captures how speed develops as time moves on. Understanding integrals is vital, as it enables calculating reverse processes from known rates of change.

Initial Conditions

Initial conditions serve as essential information that defines the state of a system at the onset of observation. For this motion-related scenario:

• At \( t = 0 \), the initial condition is \( v = 0 \). This starting point simplifies solving the integral by letting you find the constant of integration.
• It confirms consistency with real-world conditions.

Integrating without knowing where to start can deliver ambiguous results. Initial conditions help you calculate specific solutions from indefinite integrals. They're like "ground zero," ensuring that any derived formula aligns with known real-world parameters.