Problem 462 - Calculus Volume 1

Question

The velocity of a bullet from a rifle can be approximated by \(v(t)=6400 t^{2}-6505 t+2686,\) where \(t\) is seconds after the shot and \(v\) is the velocity measured in feet per second. This equation only models the velocity for the first half-second after the shot: \(0 \leq t \leq 0.5\). What is the total distance the bullet travels in 0.5 sec?

Step-by-Step Solution

Answer

The bullet travels approximately 1063.205 feet in 0.5 seconds.

Step 1: Understand the Problem

We need to find the total distance traveled by the bullet in the first half-second after it is fired. The velocity function given is \(v(t) = 6400t^2 - 6505t + 2686\) and this function is applicable for \(0 \leq t \leq 0.5\).To find the distance, we need to integrate the velocity function over this time interval.

Step 2: Set Up the Integral for Distance

The distance traveled by an object is the integral of its velocity function over a given time interval. We need to integrate \(v(t)\) from \(t=0\) to \(t=0.5\). Thus, the integral we need to compute is:\[\int_{0}^{0.5} (6400t^2 - 6505t + 2686)\,dt.\]

Step 3: Integrate the Velocity Function

The antiderivative of the velocity function \(v(t) = 6400t^2 - 6505t + 2686\) can be found by integrating each term separately.- The integral of \(6400t^2\) is \(\frac{6400}{3}t^3\)- The integral of \(-6505t\) is \(-\frac{6505}{2}t^2\)- The integral of \(2686\) is \(2686t\)Thus, the antiderivative is \[ \frac{6400}{3}t^3 - \frac{6505}{2}t^2 + 2686t. \]

Step 4: Evaluate the Integral

With the antiderivative \( \frac{6400}{3}t^3 - \frac{6505}{2}t^2 + 2686t\), we now evaluate it from \(t=0\) to \(t=0.5\).Calculate the antiderivative at \(t=0.5\) and \(t=0\):- At \(t=0.5\): \[\frac{6400}{3}(0.5)^3 - \frac{6505}{2}(0.5)^2 + 2686(0.5)\]- At \(t=0\): \[\frac{6400}{3}(0)^3 - \frac{6505}{2}(0)^2 + 2686(0) = 0.\]

Step 5: Compute the Numerical Solution

Substitute \(t=0.5\) into the antiderivative and compute the result:\[\frac{6400}{3}(0.5)^3 - \frac{6505}{2}(0.5)^2 + 2686(0.5) = 533.33 - 813.125 + 1343 = 1063.205\] (approximately).Since \(\int_{0}^{0} v(t) \, dt = 0\), the total distance traveled in 0.5 seconds is approximately 1063.205 feet.

Key Concepts

Velocity Function

When exploring how fast an object is moving, the velocity function gives us a direct measurement. In this exercise, the velocity function provided is \( v(t) = 6400t^2 - 6505t + 2686 \).
This function represents how the speed of the bullet changes over time, specifically within the first half-second after it is fired from a rifle. Velocity functions are crucial because they tell us not just the speed, but also the direction of an object.

• \( t \) is the time in seconds since the shot was fired.
• \( v(t) \) is the velocity in feet per second.

It's essential to understand that different types of motion problems, whether linear or projectile, will have unique functions representing how an object speeds up or slows down over time.

Distance Calculation

The key to finding out how far the bullet travels is calculating the distance, which involves integrating the velocity function over the given time period.
This is because the distance traveled is essentially the accumulation of all the small bits of distance covered in each time interval. In mathematical terms, this means we need to compute the definite integral of the velocity function \( v(t) \) from \( t = 0 \) to \( t = 0.5 \):\[\int_{0}^{0.5} (6400t^2 - 6505t + 2686) \, dt.\]

• This integral will sum up all the infinitesimally small distances covered, giving us the total distance.
• This comprehensive integration will help track how the rapid speed variations accumulate over time, providing an accurate representation of the total distance.

As shown, integration is indispensable in accurately measuring the distance traveled over time when the speed is variable and not constant.

Antiderivative

To solve the integral, we first need to find the antiderivative of the velocity function. An antiderivative reverses the process of differentiation, helping us understand the original function that, when differentiated, gives us the velocity.
For this specific function \( v(t) = 6400t^2 - 6505t + 2686 \), we find:

• The antiderivative of \( 6400t^2 \) is \( \frac{6400}{3}t^3 \).
• For \( -6505t \), it is \( -\frac{6505}{2}t^2 \).
• And for \( 2686 \), it is \( 2686t \).

All together, the antiderivative is:\[ \frac{6400}{3}t^3 - \frac{6505}{2}t^2 + 2686t.\]By evaluating this antiderivative from \( t = 0 \) to \( t = 0.5 \), we can determine the total distance traveled by the bullet during that time period. Antiderivatives allow us to bridge the gap from the rate of change (velocity) to the total change (distance).