Problem 72 - Calculus Single Variable

Question

A bullet is fired into a sand pile with an initial speed of 1024 feet per second. The rate at which its velocity is decreased after entering the sand pile is 500 times the \(\frac{4^{\text {th }}}{5}\) power of the velocity if units of feet and seconds are used. How long is it before the bullet comes to rest?

Step-by-Step Solution

Answer

The bullet comes to rest after \( \frac{1}{25} \) seconds.

Step 1: Understand the Velocity Change Rate

The rate at which the velocity of the bullet changes can be expressed as \( \frac{dv}{dt} = -500 \cdot v^{4/5}, \) where \( v \) is the velocity of the bullet.

Step 2: Separate Variables

To solve the differential equation, we separate variables: \( \frac{1}{v^{4/5}} \, dv = -500 \, dt. \)

Step 3: Integrate Both Sides

Integrating both sides: \( \int v^{-4/5} \, dv = -500 \int dt. \) This gives \( \frac{v^{1/5}}{1/5} = -500t + C. \)

Step 4: Simplify the Integration Result

Simplifying the integral, we have \( 5v^{1/5} = -500t + C. \)

Step 5: Find the Constant of Integration

Using the initial condition \( v = 1024 \) when \( t = 0 \), substitute to find \( C \): \( 5(1024)^{1/5} = C. \)

Step 6: Simplify the Equation with Initial Condition

Calculate \( (1024)^{1/5} = 4 \), so \( C = 5 \times 4 = 20. \) Thus, \( 5v^{1/5} = -500t + 20. \)

Step 7: Solve for t When the Bullet Comes to Rest

When the bullet comes to rest, \( v = 0. \) Substitute into the equation: \( 5(0)^{1/5} = -500t + 20, \) so \( 0 = -500t + 20. \)

Step 8: Solve the Equation for t

Re-arrange to find \( t: \)\(-500t = -20,\)\( t = \frac{-20}{-500} = \frac{1}{25}.\)

Key Concepts

Velocity Change Rate

The concept of the velocity change rate is crucial for understanding how the speed of an object varies over time, especially when influenced by external forces. In this exercise, the bullet's velocity decreases as it moves through the sand pile. The rate at which this deceleration occurs is given by a specific formula, \( \frac{dv}{dt} = -500 \cdot v^{4/5} \).
This equation tells us that the change in velocity (\( \frac{dv}{dt} \)) depends on the current velocity itself, raised to the power of \( \frac{4}{5} \).

Key aspects to consider include:

• The negative sign indicates that the velocity is decreasing, not increasing.
• The coefficient 500 represents a constant rate that determines how aggressively the bullet slows down.
• A higher initial velocity results in a faster initial rate of deceleration due to the power term \( v^{4/5} \).

Understanding this rate of change helps us to set up the appropriate differential equation for solving when the bullet will come to rest.

Separation of Variables

Separation of Variables is a classic method used to solve differential equations. By separating the variables, we can integrate each side of the equation independently.
In our case, the differential equation \( \frac{dv}{dt} = -500 \cdot v^{4/5} \) is separated by rearranging the terms:

• Move the terms involving \( v \) to one side: \( \frac{1}{v^{4/5}} \, dv \)
• Move the terms involving \( t \) to the other: \( -500 \, dt \)

Now, the equation becomes \( \frac{1}{v^{4/5}} \, dv = -500 \, dt \), making it ready for integration.
This separation allows us to handle each aspect of the equation—velocity and time—independently, simplifying the integration process.

Initial Condition

The concept of an initial condition is a key part in solving differential equations. It provides a specific value of the function you're seeking, usually at the start, to pin down the solution that fits the physical problem.
In our exercise, the initial condition specifies that the bullet's velocity \( v \) is 1024 feet per second at time \( t = 0 \).

Understanding initial conditions involves:

• Using this specific point to find the constant of integration. It's the 'C' in the integrated equation.
• Substituting this condition into the integrated equation to calculate 'C'.

In this context, we calculate \( C \) by solving the equation \( 5v^{1/5} = C \) when \( v = 1024 \). The initial condition helps to tailor the generic solution of the differential equation to fit this particular scenario.

Integration

Integration is used to solve the separated differential equation resulting from the separation of variables. It is the process of finding the antiderivative or the 'whole' from its derivative counterpart.
In our exercise:

• The velocity side of the equation is integrated as \( \int v^{-4/5} \, dv \), which yields \( \frac{v^{1/5}}{1/5} \).
• The time side is integrated straightforwardly as \( -500\int dt \), which results in \( -500t \).

The integration is then simplified to \( 5v^{1/5} = -500t + C \).
This integral incorporates the constant of integration \( C \), determined using the initial condition. Integration essentially 'uncovers' the function describing velocity as it changes over time, allowing us to solve how long it takes for the bullet to rest.