Problem 13
Question
If \(s(t)\) denotes the position of an object that moves along a straight line, then \(\Delta s / \Delta t\), called the average velocity, is the average rate of change of \(s(t)\), and \(v(t)=d s / d t\), called the (instantaneous) velocity, is the instantaneous rate of change of \(s(t) .\) The speed of the object is the absolute value of the velocity, \(|v(t)|\). Suppose now that a car moves along a straight road. The location at time \(t\) is given by $$ s(t)=\frac{160}{3} t^{2}, \quad 0 \leq t \leq 1 $$ where \(t\) is measured in hours and \(s(t)\) is measured in kilometers. (a) Where is the car at \(t=3 / 4\), and where is it at \(t=1 ?\) (b) Find the average velocity of the car between \(t=3 / 4\) and \(t=1\) (c) Find the velocity and the speed of the car at \(t=3 / 4\).
Step-by-Step Solution
Verified Answer
At \( t=3/4 \), the car is at 30 km, and at \( t=1 \), it is at 53.33 km. The average velocity is 93.32 km/h. The velocity and speed at \( t=3/4 \) are 80 km/h.
1Step 1: Find Position at Specific Times
To determine where the car is at specific times, we will substitute values of \( t \) into the position function \( s(t) = \frac{160}{3} t^2 \). For \( t = \frac{3}{4} \), \( s\left(\frac{3}{4}\right) = \frac{160}{3} \left(\frac{3}{4}\right)^2 = \frac{160}{3} \times \frac{9}{16} = \frac{160 \times 9}{3 \times 16} = 30 \) kilometers. For \( t = 1 \), \( s(1) = \frac{160}{3} \times 1^2 = \frac{160}{3} \approx 53.33 \) kilometers.
2Step 2: Calculate Average Velocity
The average velocity \( \frac{\Delta s}{\Delta t} \) is defined as the change in position divided by the change in time. Thus, between \( t = \frac{3}{4} \) and \( t = 1 \), the average velocity is \( \frac{s(1) - s\left(\frac{3}{4}\right)}{1 - \frac{3}{4}} = \frac{\frac{160}{3} - 30}{\frac{1}{4}} = \frac{53.33 - 30}{0.25} = \frac{23.33}{0.25} = 93.32 \) km/h.
3Step 3: Differentiate Position to Find Velocity
To find the instantaneous velocity \( v(t) \), take the derivative of the position function \( s(t) = \frac{160}{3} t^2 \). Thus, \( v(t) = \frac{d}{dt} \left(\frac{160}{3} t^2\right) = \frac{160}{3} \times 2t = \frac{320}{3} t \).
4Step 4: Evaluate Velocity and Speed at Specific Time
Substitute \( t = \frac{3}{4} \) into the velocity function: \( v\left(\frac{3}{4}\right) = \frac{320}{3} \times \frac{3}{4} = 80 \) km/h. Since speed is the absolute value of velocity, the speed at \( t = \frac{3}{4} \) is also \( 80 \) km/h.
Key Concepts
Understanding the Position FunctionDiscovering Instantaneous VelocityThe Role of Calculus in Motion Analysis
Understanding the Position Function
In physics, the position function is crucial for determining an object's location over time. It's a mathematical function that describes the location of an object along a straight line at any given moment. In our exercise, the position function is defined as \( s(t) = \frac{160}{3} t^2 \). This formula shows us how far the car has traveled at time \( t \) hours.
To pinpoint the car's location at a particular time, we substitute the desired time value into the position function. For example, to find where the car is at \( t = \frac{3}{4} \), we'd compute \( s\left(\frac{3}{4}\right) = \frac{160}{3} \times \left(\frac{3}{4}\right)^2 \). Solving this at the specified time gives us the car's precise position in kilometers.
To pinpoint the car's location at a particular time, we substitute the desired time value into the position function. For example, to find where the car is at \( t = \frac{3}{4} \), we'd compute \( s\left(\frac{3}{4}\right) = \frac{160}{3} \times \left(\frac{3}{4}\right)^2 \). Solving this at the specified time gives us the car's precise position in kilometers.
Discovering Instantaneous Velocity
Instantaneous velocity refers to the speed and direction of an object at a specific moment. It's different from average velocity, which measures speed over a period of time. To find the instantaneous velocity, we need to look at the derivative of the position function.
The given position function is \( s(t) = \frac{160}{3} t^2 \). By differentiating this function, you find \( v(t) = \frac{d}{dt} \left(\frac{160}{3} t^2\right) = \frac{320}{3} t \). This derivative represents the car's velocity at any time \( t \).
To find the instantaneous velocity at \( t = \frac{3}{4} \), substitute the value into the velocity function: \( v\left(\frac{3}{4}\right) = \frac{320}{3} \times \frac{3}{4} = 80 \text{ km/h} \).
This calculation shows how fast and in what direction the car is moving at that exact moment.
The given position function is \( s(t) = \frac{160}{3} t^2 \). By differentiating this function, you find \( v(t) = \frac{d}{dt} \left(\frac{160}{3} t^2\right) = \frac{320}{3} t \). This derivative represents the car's velocity at any time \( t \).
To find the instantaneous velocity at \( t = \frac{3}{4} \), substitute the value into the velocity function: \( v\left(\frac{3}{4}\right) = \frac{320}{3} \times \frac{3}{4} = 80 \text{ km/h} \).
This calculation shows how fast and in what direction the car is moving at that exact moment.
The Role of Calculus in Motion Analysis
Calculus is essential in analyzing motion because it allows us to understand changes in position and speed. The position and velocity of an object are not always constant; rather, they frequently vary over time. Calculus provides the tools to explore these variations.
By employing derivatives, calculus lets us find instantaneous rates of change, like velocity, from position functions. Conversely, integration can give us the total distance traveled from a velocity function.
In this exercise, taking the derivative of the position function \( s(t) = \frac{160}{3} t^2 \) yielded the velocity function \( v(t) = \frac{320}{3} t \), demonstrating how calculus helps us transition from knowing where an object is to determining how fast it's moving at any instant.
By employing derivatives, calculus lets us find instantaneous rates of change, like velocity, from position functions. Conversely, integration can give us the total distance traveled from a velocity function.
In this exercise, taking the derivative of the position function \( s(t) = \frac{160}{3} t^2 \) yielded the velocity function \( v(t) = \frac{320}{3} t \), demonstrating how calculus helps us transition from knowing where an object is to determining how fast it's moving at any instant.
Other exercises in this chapter
Problem 13
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