Problem 10 - Calculus Early Transcendentals

Question

Assume \(t\) is time measured in seconds and velocities have units of \(\mathrm{m} / \mathrm{s}\). a. Graph the velocity function over the given interval. Then determine when the motion is in the positive direction and when it is in the negative direction. b. Find the displacement over the given interval. c. Find the distance traveled over the given interval. $$v(t)=-t^{2}+5 t-4 ; 0 \leq t \leq 5$$

Step-by-Step Solution

Answer

Answer: The motion is in the positive direction in the intervals \(0 \leq t < 1\) and \(4 < t \leq 5\), and in the negative direction in the interval \(1 < t < 4\). The displacement of the function is \(\frac{8}{3}\) meters, and the distance traveled is \(6\) meters.

Step 1: Identify the critical points of the velocity function

In order to find the critical points of the velocity function, set \(v(t)=0\) and then solve for \(t\). $$-t^2 + 5t - 4 = 0$$ By solving this quadratic equation, either using factoring or quadratic formula, we find the critical points to be: $$t = 1, t = 4$$

Step 2: Test intervals for positive and negative motion

Now we'll analyze the intervals around the critical points to determine when the motion is positive and negative: 1. Interval \(0 \leq t < 1\) 2. Interval \(1 < t < 4\) 3. Interval \(4 < t \leq 5\) We can test each of these intervals by plugging in a test point from each interval into the velocity function, and then checking if the value of the function is positive or negative: 1. For example, when \(t=0.5\), \(v(0.5)=-0.5^2+5(0.5)-4 > 0\), so the motion is in the positive direction. 2. With \(t=2\), \(v(2)= -2^2+5(2)-4 < 0\), so the motion is in the negative direction. 3. And for \(t=4.5\), \(v(4.5)=-4.5^2+5(4.5)-4 > 0\), so the motion is in the positive direction. Now we can graph the function and label the positive and negative regions accordingly. b. Find the displacement over the given interval

Step 3: Integrate the velocity function to find displacement

To find the displacement, we need to integrate the velocity function between \(t=0\) and \(t=5\). $$\int_{0}^{5} (-t^{2} + 5t - 4) \, dt$$ Applying the integral to each term, we get: $$\left[-\frac{1}{3}t^{3} +\frac{5}{2}t^{2} - 4t\right]_{0}^{5}$$ Now evaluate the expression at the bounds and take the difference: $$\left[-\frac{1}{3}(5^{3}) +\frac{5}{2}(5^{2}) - 4(5)\right]-\left[-\frac{1}{3}(0^{3}) +\frac{5}{2}(0^{2}) - 4(0)\right]$$ Simplify the expression to find the displacement: $$displacement = -\frac{1}{3}(125) +\frac{5}{2}(25) - 20 = \frac{8}{3} \mathrm{m}$$ c. Find the distance traveled over the given interval

Step 4: Find the net distance traveled by taking the sum of the absolute values of positive/negative regions

Since we already know the intervals in which motion was positive or negative, we can find the distance traveled by integrating the absolute value of the velocity function separately for each interval and summing their results: Positive motion intervals are \(0 \leq t < 1\) and \(4 < t \leq 5\) and the negative motion interval is \(1 < t < 4\). $$distance = \int_{0}^{1} v(t) \, dt - \int_{1}^{4} v(t) \, dt + \int_{4}^{5} v(t) \, dt$$ Now we integrate the function and take the sum of the absolute values of the integrals in each interval: $$distance = \left[-\frac{1}{3}t^{3} +\frac{5}{2}t^{2} - 4t\right]_{0}^{1} - \left[-\frac{1}{3}t^{3} +\frac{5}{2}t^{2} - 4t\right]_{1}^{4} + \left[-\frac{1}{3}t^{3} +\frac{5}{2}t^{2} - 4t\right]_{4}^{5}$$ After calculating, we find $$distance = \frac{18}{3} \mathrm{m}$$

Key Concepts

Velocity

Velocity is a key concept in calculus and physics and is defined as the rate of change of an object's position over time. It tells us how fast an object is moving and in which direction. In our exercise, the velocity function is given by:\[-t^2 + 5t - 4\]Here, velocity is a function of time \(t\), measured in seconds, and the unit of velocity is meters per second (m/s). This function is a quadratic expression, indicating that the velocity changes in a non-linear manner as time progresses. It's important to recognize that velocity can be both positive and negative:

• Positive velocity means the object is moving forward or in a positive direction.
• Negative velocity indicates motion in the opposite direction or backward.

In our step-by-step solution, by setting the velocity function \(v(t)\) to zero, we find the critical points (\(t = 1\) and \(t = 4\)), which help us understand when the velocity changes direction.

Displacement

Displacement in calculus refers to the change in position of an object in a particular direction between two points of time. It is a vector quantity, meaning it has both magnitude and direction. In the exercise, displacement is found by integrating the velocity from the start time, \(t_0\), to the end time, \(t_f\). This integration finds both shape and area under the velocity-time graph, representing the net change in position. For the given velocity function, the displacement over the interval \(0 \leq t \leq 5\) is calculated by the integral:\[\int_{0}^{5} (-t^{2} + 5t - 4) \, dt\]The displacement calculated is \(\frac{8}{3} \text{ meters}\). This quantity, though taking negative and positive velocities into account, tells us the overall change in position. So, even if parts of the journey were in opposite directions, only the resultant position shift is presented.

Distance

Distance differs from displacement because it accounts for the entire path traveled, regardless of direction. In calculus, the distance traveled is found by integrating the absolute value of the velocity function over the desired interval. Why absolute value? Because distance cannot be negative and it accumulates all travel paths the object follows. To compute the distance traveled in our exercise, we consider the segments where velocity changes are accounted by splitting the integral when the direction changes, for the intervals:\

• \(0 \leq t < 1\)
• \(1 < t < 4\)
• \(4 < t \leq 5\)

The overall distance is comprised of positive contributions from all parts of the journey, leading us to the solution \(\frac{18}{3} \text{ meters}\). This total considers the complete journey path, ignoring direction changes.

Quadratic Equation

The quadratic equation is fundamental in many fields, including calculus, where it often describes acceleration and velocity functions. A quadratic equation is generally expressed in the form \(ax^2 + bx + c = 0\). It forms a parabola when graphed. In the given exercise, the velocity function \(-t^2 + 5t - 4\) serves as a quadratic equation:

• The coefficient \(-1\) (for \(t^2\)) shows that the parabola opens downwards.
• The roots of the equation, found by solving \(-t^2 + 5t - 4 = 0\), give the times when the velocity is zero (critical points).
• These roots are calculated using the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• In our exercise, the roots play a crucial role in determining the intervals for positive and negative velocity.

Understanding quadratic equations helps interpret the behavior of the velocity over time, significantly influencing displacement and distance calculations.