Problem 47 - Calculus and its applications

Question

Given $$ s(t)=3 t+10 $$ where \(s(t)\) is in miles and \(t\) is in hours, find each of the following. a) \(v(t)\) b) \(a(t)\) c) The velocity and acceleration when \(t=2 \mathrm{hr}\) d) When the distance function is given by a linear function, we have uniform motion. What does uniform motion mean in terms of velocity and acceleration?

Step-by-Step Solution

Answer

a) \(v(t) = 3\) mph; b) \(a(t) = 0\) mph²; c) Velocity: 3 mph, Acceleration: 0 mph²; d) Uniform motion: constant velocity, zero acceleration.

Step 1: Calculate the Velocity Function

The velocity function, \(v(t)\), can be obtained by differentiating the position function \(s(t)\) with respect to time \(t\). The position function is \(s(t) = 3t + 10\). Differentiating with respect to \(t\), we get:\[ v(t) = \frac{d}{dt}[3t + 10] = 3. \] Thus, the velocity function is constant: \(v(t) = 3\, \text{miles per hour}.\)

Step 2: Calculate the Acceleration Function

The acceleration function \(a(t)\) can be found by differentiating the velocity function \(v(t)\) with respect to \(t\). Since \(v(t) = 3\) is a constant, its derivative is zero:\[ a(t) = \frac{d}{dt}[3] = 0. \] Therefore, the acceleration function is \(a(t) = 0\, \text{miles per hour squared}.\)

Step 3: Evaluate Velocity and Acceleration at \(t = 2\) Hours

We need to find the values of \(v(t)\) and \(a(t)\) when \(t = 2\). Since both \(v(t)\) and \(a(t)\) are constants, their values at any \(t\) are:\[ v(2) = 3\, \text{miles per hour}, \] and \[ a(2) = 0\, \text{miles per hour squared}. \] Therefore, the velocity is \(3\, \text{miles per hour}\) and the acceleration is \(0\, \text{miles per hour squared}\) at \(t = 2\).

Step 4: Explain Uniform Motion

Uniform motion occurs when an object moves at a constant velocity. This means that the velocity does not change over time, and hence, the acceleration is zero. In terms of this problem, since the distance function \(s(t)\) is linear, it implies the velocity is constant (\(v(t) = 3\)), and the acceleration is zero (\(a(t) = 0\)).

Key Concepts

Velocity in Calculus

Velocity in calculus is a key concept used to describe how fast an object moves along a path over time. When we talk about velocity, we're referring to a vector quantity that not only includes the speed of an object but also its direction. Simply put, velocity tells us how the position of an object changes with time.

To find velocity mathematically, we use differentiation. Differentiation is a process that calculates the derivative of a function. This derivative gives us the rate at which the function's value is changing at any point.

In the given problem, the position function is defined as:

• \( s(t) = 3t + 10 \), where \( s(t) \) represents the position of an object, and \( t \) is time in hours.

By differentiating \( s(t) \) with respect to \( t \), we obtain the velocity function \( v(t) \).

• \( v(t) = \frac{d}{dt}[3t + 10] = 3 \).

This result tells us that the object moves at a constant velocity of 3 miles per hour.

Understanding Acceleration

Acceleration is another fundamental concept in calculus, described as the rate of change of velocity with respect to time. In simpler terms, acceleration tells us how quickly an object speeds up or slows down. It is a vector quantity, just like velocity.

In the problem at hand, we derived that the velocity function is a constant \( v(t) = 3 \). To find the acceleration, we differentiate the velocity function:

• \( a(t) = \frac{d}{dt}[3] = 0 \).

The acceleration function \( a(t) = 0 \) miles per hour squared indicates that there is no change in velocity; the object doesn't speed up or slow down as time progresses. This makes the motion uniform.

The Role of Differentiation

Differentiation is a core technique in calculus that provides tools for understanding rates of change. It is essential for finding how quantities increase or decrease over time, such as velocity and acceleration.

In the context of physics and motion, differentiation allows us to find:

• The velocity by differentiating position functions.
• The acceleration by differentiating velocity functions.

For the exercise, the position function \( s(t) = 3t + 10 \) was differentiated to find a constant velocity \( v(t) = 3 \) and a zero acceleration \( a(t) = 0 \). Using differentiation effectively, we can analyze and predict the movement of objects.

Understanding Uniform Motion

Uniform motion is a straightforward yet vital concept in physics and calculus. It refers to the movement of an object at a constant velocity, meaning there is no change in its speed or direction over time. Uniform motion implies that acceleration is zero.

In the given problem, the linear position function \( s(t) = 3t + 10 \) illustrates uniform motion. Why? Because the velocity \( v(t) \) derived from it is a constant 3 miles per hour, and the acceleration \( a(t) = 0 \) indicates no change in speed.

Uniform motion example applications include vehicles moving at a steady speed on a straight road or a conveyor belt maintaining a constant pace. Recognizing uniform motion helps in real-world scenarios for predicting behavior and outcomes based on constant movement.