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Magazine Chapter 6: Problem 35
Starting at \(s=0\) when \(t=0\), an object moves along a line so that its velocity at time \(t\) is \(v(t)=2 t-4\) centimeters per second. How long will it take to get to \(s=12\) ? To travel a total distance of 12 centimeters?
Short Answer
Expert verified
The object takes 6 seconds to travel a total distance of 12 cm.
Step by step solution
01
Understand the Problem
The problem involves finding the time it takes for an object to travel 12 cm given its velocity as a function of time, \(v(t) = 2t - 4\, \text{cm/s}\), starting at \(s = 0\, \text{cm}\) when \(t = 0\). We need to find the time \(t\) when \(s = 12\, \text{cm}\).
02
Set Up the Position Function
The velocity function \(v(t) = 2t - 4\) describes the rate of change of position \(s(t)\). To find \(s(t)\), we integrate \(v(t)\) with respect to \(t\): \[ s(t) = \int (2t - 4) \, dt = t^2 - 4t + C \], where \(C\) is the constant of integration. Since \(s(0) = 0\), we substitute: \(0 = 0^2 - 4 \times 0 + C\), giving \(C = 0\). So, \(s(t) = t^2 - 4t\).
03
Solve for Time When Displacement Equals 12 cm
Now, we need \(s(t) = 12\, \text{cm}\). Set up the equation:\[ t^2 - 4t = 12. \]Rearrange to form a quadratic equation:\[ t^2 - 4t - 12 = 0. \]Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \(a = 1\), \(b = -4\), and \(c = -12\).
04
Calculate the Positive Time Solution
Substitute \(a = 1\), \(b = -4\), \(c = -12\) into the quadratic formula:\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-12)}}{2 \times 1}. \]\[ t = \frac{4 \pm \sqrt{16 + 48}}{2}. \]\[ t = \frac{4 \pm \sqrt{64}}{2}. \]\[ t = \frac{4 \pm 8}{2}. \]This gives \( t = 6 \) or \(-2\). Since time cannot be negative, \( t = 6 \).
05
Verify Completing Travel for Total Distance of 12 cm
Check if the object changes direction before reaching 12 cm by finding when \(v(t) = 0\):\[ 2t - 4 = 0 \Rightarrow t = 2. \]For \(0 \leq t \leq 2\), the object moves forward. Calculate displacement:\(s(2) = 2^2 - 4 \times 2 = -4\, \text{cm}\), indicating a reversal. From \(t = 2\) to \(t = 6\), it moves backward from \(s=0\) to \(s=12\). So, it indeed travels a total of 12 cm.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity Function
The velocity function tells us how fast an object's position is changing over time. In this exercise, the velocity is given by the function \( v(t) = 2t - 4 \), where \( t \) is the time in seconds and \( v(t) \) is the velocity in centimeters per second.
This means at any time \( t \), the velocity of the object can be found by plugging \( t \) into this equation.
Understanding this function helps us predict how quickly the object is moving and in which direction it is going.
This means at any time \( t \), the velocity of the object can be found by plugging \( t \) into this equation.
Understanding this function helps us predict how quickly the object is moving and in which direction it is going.
- At \( t = 0 \): Velocity is \( v(0) = 2 \times 0 - 4 = -4 \) cm/s, indicating backward movement.
- At \( t = 2 \): Velocity becomes \( v(2) = 2 \times 2 - 4 = 0 \) cm/s, meaning the object momentarily stops before changing direction.
- At \( t = 6 \): Velocity is \( v(6) = 2 \times 6 - 4 = 8 \) cm/s, representing forward movement.
Integration
Integration is a fundamental concept in calculus used to find the position function from a given velocity function. When we integrate a velocity function, we essentially accumulate all the tiny changes in position over the given time interval. In this particular exercise, we integrated \( v(t) = 2t - 4 \) to find the position function \( s(t) \).
The process of integration can be visualized as adding up velocities over time to discover the total change in position.
Let's break it down:
The process of integration can be visualized as adding up velocities over time to discover the total change in position.
Let's break it down:
- Step 1: Set up the integral: \( s(t) = \int (2t - 4) \, dt \)
- Step 2: Solve the integral: \( s(t) = t^2 - 4t + C \)
- Step 3: Apply initial conditions to find the constant \( C \).
Quadratic Equation
A quadratic equation is an equation where the highest power of the variable is 2. In this exercise, after setting up the equation for position, we ended up with \( t^2 - 4t = 12 \) which rearranges to the quadratic equation \( t^2 - 4t - 12 = 0 \).
This form is typical in calculus for describing paths and trajectories.
This form is typical in calculus for describing paths and trajectories.
- Standard Form: \( ax^2 + bx + c = 0 \)
- Solution Method: We used the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Discriminant: \( b^2 - 4ac \) determines the nature of the roots (real and distinct, real and same, or complex).
Constant of Integration
The constant of integration, represented as \( C \), is crucial when determining an indefinite integral. After integrating a velocity function like \( v(t) = 2t - 4 \), you obtain a general solution \( s(t) = t^2 - 4t + C \).
However, this \( C \) can take an infinite number of values until you use initial conditions to find its exact value.
However, this \( C \) can take an infinite number of values until you use initial conditions to find its exact value.
- Purpose: Accounts for any vertical shift in the position function.
- Finding \( C \): By using given conditions, like \( s(0) = 0 \), we find \( C = 0 \).
- Effect: Ensures that the position function correctly models the object's motion from the specified starting point.
