Problem 17
Question
An object is moving along a coordinate line subject to the indicated acceleration a (in centimeters per second per second) with the initial velocity \(v_{0}\) (in centimeters per second) and directed distance \(s_{0}\) (in centimeters). Find both the velocity \(v\) and directed distance s after 2 seconds (see Example 4). \(a=t ; v_{0}=3, s_{0}=0\)
Step-by-Step Solution
Verified Answer
Velocity is 5 cm/s and position is 6.33 cm after 2 seconds.
1Step 1: Understand the Given Information
We are given an acceleration function \(a = t\), initial velocity \(v_0 = 3\, \text{cm/s}\), and initial position \(s_0 = 0\, \text{cm}\). We need to find the velocity \(v\) and position \(s\) at \(t = 2\) seconds.
2Step 2: Find the Velocity Function
To find the velocity function, we integrate the acceleration function \(a = t\) with respect to \(t\). The integral of \(t\) is \(\frac{1}{2}t^2\). Adding the constant of integration \(C\) to account for the initial velocity, we get \(v(t) = \frac{1}{2}t^2 + C\).
3Step 3: Apply Initial Condition to Velocity
Use the initial condition \(v_0 = 3\) when \(t = 0\) to determine \(C\). Thus, \(3 = \frac{1}{2}(0)^2 + C\) which gives \(C = 3\). Therefore, the velocity function is \(v(t) = \frac{1}{2}t^2 + 3\).
4Step 4: Calculate Velocity at \(t = 2\)
Substitute \(t = 2\) into the velocity equation: \(v(2) = \frac{1}{2}(2)^2 + 3 = \frac{1}{2} \times 4 + 3 = 2 + 3 = 5\, \text{cm/s}\).
5Step 5: Find the Position Function
To find the position function \(s(t)\), we integrate the velocity function \(v(t) = \frac{1}{2}t^2 + 3\) with respect to \(t\). The integral is \(\frac{1}{6}t^3 + 3t + C\).
6Step 6: Apply Initial Condition to Position
Use the initial condition \(s_0 = 0\) when \(t = 0\) to determine \(C\). Thus, \(0 = \frac{1}{6}(0)^3 + 3(0) + C\), resulting in \(C = 0\). Therefore, the position function is \(s(t) = \frac{1}{6}t^3 + 3t\).
7Step 7: Calculate Position at \(t = 2\)
Substitute \(t = 2\) into the position equation: \(s(2) = \frac{1}{6}(2)^3 + 3(2) = \frac{1}{6} \times 8 + 6 = \frac{8}{6} + 6 = \frac{4}{3} + 6 = 6.33\, \text{cm}\).
Key Concepts
Acceleration and VelocityIntegrationInitial ConditionsPosition Function
Acceleration and Velocity
Understanding the relationship between acceleration and velocity is crucial in solving many calculus problems. Acceleration tells us how fast the rate of velocity is changing over time. In this example, the acceleration function is given as \(a = t\). This indicates that as time progresses, the acceleration increases linearly.
When we know the acceleration as a function of time, we can find the velocity by integrating this function. This involves calculating the antiderivative, which essentially gives us a formula for velocity that tells us how fast the object is moving at any given time.
In our example, integrating the acceleration function \(a = t\) gives us the velocity function \(v(t) = \frac{1}{2}t^2 + C\), where \(C\) is a constant determined by initial conditions.
When we know the acceleration as a function of time, we can find the velocity by integrating this function. This involves calculating the antiderivative, which essentially gives us a formula for velocity that tells us how fast the object is moving at any given time.
In our example, integrating the acceleration function \(a = t\) gives us the velocity function \(v(t) = \frac{1}{2}t^2 + C\), where \(C\) is a constant determined by initial conditions.
Integration
Integration is a fundamental concept in calculus used to find solutions to problems involving rates of change, such as acceleration and velocity. It reverses differentiation by finding the original function given its rate of change.
In the example provided, the acceleration \(a = t\) is integrated with respect to time \(t\) to find the velocity function. This involves calculating the antiderivative, resulting in \(\int t \, dt = \frac{1}{2}t^2 + C\).
Similarly, integrating the velocity function \(v(t) = \frac{1}{2}t^2 + 3\) determines the position function, \(s(t) = \frac{1}{6}t^3 + 3t + C\), giving us the object’s location over time.
In the example provided, the acceleration \(a = t\) is integrated with respect to time \(t\) to find the velocity function. This involves calculating the antiderivative, resulting in \(\int t \, dt = \frac{1}{2}t^2 + C\).
Similarly, integrating the velocity function \(v(t) = \frac{1}{2}t^2 + 3\) determines the position function, \(s(t) = \frac{1}{6}t^3 + 3t + C\), giving us the object’s location over time.
Initial Conditions
Initial conditions provide vital information that allows us to solve differential equations with integration. They specify the value of a function or its derivatives at a particular point in time, helping to determine the unknown constant \(C\) in the integral.
In this problem, the initial velocity \(v_0 = 3\, \text{cm/s}\) is used to calculate \(C\) in the velocity function \(v(t) = \frac{1}{2}t^2 + C\). By substituting \(t = 0\), we derive \(C = 3\).
Similarly, the initial position \(s_0 = 0\, \text{cm}\) is used when calculating the position function's constant, resulting in the position function \(s(t) = \frac{1}{6}t^3 + 3t\).
In this problem, the initial velocity \(v_0 = 3\, \text{cm/s}\) is used to calculate \(C\) in the velocity function \(v(t) = \frac{1}{2}t^2 + C\). By substituting \(t = 0\), we derive \(C = 3\).
Similarly, the initial position \(s_0 = 0\, \text{cm}\) is used when calculating the position function's constant, resulting in the position function \(s(t) = \frac{1}{6}t^3 + 3t\).
Position Function
The position function provides a way to track where an object is located in space over time. It's derived by integrating the velocity function, which itself is an antiderivative of the acceleration.
In our example, the position function is \(s(t) = \frac{1}{6}t^3 + 3t\). This gives us the location of the object in centimeters at any point in time \(t\). Determining such a function is essential for understanding an object's journey from one point to another.
To find the precise position after 2 seconds, substitute \(t = 2\) into \(s(t)\), yielding \(s(2) = 6.33\, \text{cm}\). This tells us the object's directed distance from the start location at that specific time.
In our example, the position function is \(s(t) = \frac{1}{6}t^3 + 3t\). This gives us the location of the object in centimeters at any point in time \(t\). Determining such a function is essential for understanding an object's journey from one point to another.
To find the precise position after 2 seconds, substitute \(t = 2\) into \(s(t)\), yielding \(s(2) = 6.33\, \text{cm}\). This tells us the object's directed distance from the start location at that specific time.
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