Problem 33 - Calculus Early Transcendentals

Question

A drag racer accelerates at \(a(t)=88 \mathrm{ft} / \mathrm{s}^{2} .\) Assume that \(v(0)=0\) and \(s(0)=0\). a. Determine and graph the position function, for \(t \geq 0\). b. How far does the racer travel in the first 4 seconds? c. At this rate, how long will it take the racer to travel \(\frac{1}{4}\) mi? d. How long does it take the racer to travel \(300 \mathrm{ft} ?\) e. How far has the racer traveled when it reaches a speed of \(178 \mathrm{ft} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

a. The position function is \(s(t) = 44t^2\), and its graph is a parabola with the equation \(y = 44x^2\) for \(x \geq 0\). b. The drag racer travels 704 feet in the first 4 seconds. c. It takes the racer \(\sqrt{30}\) seconds to travel \(\frac{1}{4}\) mi. d. It takes the racer \(\sqrt{\frac{300}{44}}\) seconds to travel 300 feet. e. The racer has traveled \(\frac{39663}{22}\) feet when it reaches a speed of 178 ft/s.

Step 1: Integrate the acceleration function to find the velocity function

First, we need to find the velocity function by integrating the acceleration function with respect to time (t): $$ v(t) = \int a(t) \, dt = \int 88 \, dt $$ Now, integrate the function: $$ v(t) = 88t + C_1 $$ Given that \(v(0)=0\), we can find the value of constant \(C_1\) by substituting \(t=0\) and \(v(0)=0\): $$ 0 = 88(0) + C_1 \implies C_1 = 0 $$ So, the velocity function is: $$ v(t) = 88t $$

Step 2: Integrate the velocity function to find the position function

Next, we need to find the position function by integrating the velocity function with respect to time (t): $$ s(t) = \int v(t) \, dt = \int 88t \, dt $$ Now, integrate the function: $$ s(t) = 44t^2 + C_2 $$ Given that \(s(0)=0\), we can find the value of constant \(C_2\) by substituting \(t=0\) and \(s(0)=0\): $$ 0 = 44(0)^2 + C_2 \implies C_2 = 0 $$ So, the position function is: $$ s(t) = 44t^2 $$ This is the position function for \(t \geq 0\). To graph this function, plot the parabola \(y=44x^2\) for \(x \geq 0\).

Step 3: Find the distance traveled in the first 4 seconds

Substitute \(t=4\) in the position function \(s(t)\): $$ s(4) = 44(4)^2 = 44(16) = 704 \, \text{ft} $$ So, the racer travels 704 feet in the first 4 seconds.

Step 4: Find the time taken to travel \(\frac{1}{4}\) mi

First, we need to convert \(\frac{1}{4}\) mi to feet. There are 5280 feet in a mile, so \(\frac{1}{4}\) mi is equivalent to \(\frac{1}{4} \times 5280 = 1320\) feet. We want to find the value of \(t\) such that \(s(t) = 1320\): $$ 44t^2 = 1320 $$ Divide both sides by 44: $$ t^2 = 30 $$ Take the square root of both sides and keep only the positive value since \(t \geq 0\): $$ t = \sqrt{30} $$ So, it takes the racer \(\sqrt{30}\) seconds to travel \(\frac{1}{4}\) mi.

Step 5: Find the time taken to travel 300 ft

We want to find the value of \(t\) such that \(s(t) = 300\): $$ 44t^2 = 300 $$ Divide both sides by 44: $$ t^2 = \frac{300}{44} $$ Take the square root of both sides and keep only the positive value since \(t \geq 0\): $$ t = \sqrt{\frac{300}{44}} $$ So, it takes the racer \(\sqrt{\frac{300}{44}}\) seconds to travel 300 feet.

Step 6: Find the distance traveled when speed is 178 ft/s

We want to find the value of \(t\) such that \(v(t) = 178\): $$ 88t = 178 $$ Divide both sides by 88: $$ t = \frac{178}{88} $$ Now, substitute this value of \(t\) in the position function \(s(t)\): $$ s\left(\frac{178}{88}\right) = 44\left(\frac{178}{88}\right)^2 = \frac{39663}{22} \, \text{ft} $$ So, the racer has traveled \(\frac{39663}{22}\) feet when it reaches a speed of 178 ft/s.

Key Concepts

Position Function

The position function, typically denoted by \( s(t) \), describes the location of an object at any given time \( t \). It is fundamentally important in kinematics, where we study the motion of objects. For a drag racer with its initial position set at \( s(0) = 0 \), determining its position over time involves integrating its velocity function. When we integrate velocity, which is the rate of change of position, we get the position function.

In the given problem, the velocity function is \( v(t) = 88t \), obtained by integrating the acceleration function \( a(t) = 88 \text{ ft/s}^2 \). To find the position function, we integrate the velocity function:

• Start with \( s(t) = \int 88t \, dt \).
• The integration yields \( s(t) = 44t^2 + C \).
• Using the initial condition \( s(0) = 0 \), solve for \( C \), which results in \( C = 0 \).

Thus, the position function simplifies to \( s(t) = 44t^2 \). This quadratic function indicates that the position changes with the square of time, showing a parabolic upward trajectory when graphed for \( t \geq 0 \).

Velocity Function

The velocity function \( v(t) \) is a critical component in understanding motion, representing the rate of change of position per unit time. In kinematics, it is obtained by integrating the acceleration function, which tells us how speed changes over time.

In this problem, the initial velocity is \( v(0) = 0 \) because the drag racer starts from rest. Given the acceleration function \( a(t) = 88 \text{ ft/s}^{2} \), we integrate to find the velocity:

• Perform \( v(t) = \int 88 \, dt \).
• The result is \( v(t) = 88t + C \).
• With \( v(0) = 0 \), set \( 88(0) + C = 0 \) to find \( C = 0 \).

This leads to the velocity function \( v(t) = 88t \). The linear representation shows that velocity increases constantly with time due to uniform acceleration.

Acceleration Function

The acceleration function \( a(t) \) is a key concept that describes how the velocity of an object changes over time. It is defined as the derivative of the velocity function, and integration of acceleration gives us the velocity.

In this exercise, the acceleration is given as a constant \( a(t) = 88 \text{ ft/s}^2 \). This means the drag racer experiences a constant increase in speed every second, implying a constant linear change in velocity. The motion characteristic of such a constant acceleration leads to simplified calculations for both velocity and position, as both are determined by straightforward integrations.

A constant acceleration like this one translates into simple, predictable motion patterns, which are often used in initial physics problems to teach concepts of integration and initial conditions.

Integration

Integration is a mathematical process used to calculate areas under curves, among other things, and is a fundamental operation in calculus. In kinematics, it is used to determine velocity from acceleration or position from velocity.

In this context, we utilized integration twice:

• First, to find the velocity function \( v(t) \) by integrating the constant acceleration: \( v(t) = \int a(t) \, dt \).
• Second, to find the position function \( s(t) \) by integrating the velocity function: \( s(t) = \int v(t) \, dt \).

Both steps required careful handling of constants of integration, \( C_1 \) and \( C_2 \), to be determined by given initial conditions. This process showcases how integration allows us to move backwards from rate of change to the quantity itself, and is fundamental in solving differential equations related to motion.

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the causes of this motion. It focuses on concepts such as displacement, velocity, and acceleration.

When solving problems in kinematics, like the drag racer example, three primary quantities are typically analyzed:

• **Position:** Given by \( s(t) \), it tells us where the object is in space at any time \( t \).
• **Velocity:** Described by \( v(t) \), it indicates how fast and in what direction an object is moving.
• **Acceleration:** Denoted by \( a(t) \), it reveals how the velocity of the object changes over time.

Integration is the tool kinematics relies on to link these concepts: starting from acceleration we find velocity, and from velocity, we find position. This builds a comprehensive picture of the motion, crucial for tackling real-world problems in engineering and physics.