Problem 5 - Calculus: Early Transcendentals

Question

An object moves with velocity \(v(t)=-t^{2}+1\) feet per second between \(t=0\) and \(t=2\). Find the average velocity and the average speed of the object between \(t=0\) and \(t=2\).

Step-by-Step Solution

Answer

The average velocity is \\(-\frac{1}{3}\) ft/s, and the average speed is 0.5 ft/s.

Step 1: Understand the Function

We are given the velocity function for the object: \(v(t) = -t^2 +1\). Velocity can be negative, which indicates the object is moving in the opposite direction.

Step 2: Calculate Total Displacement

Integrate the velocity function from \(t=0\) to \(t=2\) to find the displacement.\[\text{Displacement} = \int_{0}^{2} (-t^2 + 1) \, dt\]This calculates to:\[\int (-t^2 + 1) \, dt = \left[ -\frac{t^3}{3} + t \right]_{0}^{2} = \left(-\frac{8}{3} + 2 \right) - \left(0\right) = \frac{-2}{3}\]

Step 3: Calculate Average Velocity

Average velocity is calculated by dividing the total displacement by the total time:\[\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Time Interval}} = \frac{\frac{-2}{3}}{2} = -\frac{1}{3} \text{ ft/s}\]

Step 4: Calculate Total Distance

Total distance accounts for all motion regardless of direction, calculated as:The speed function is \(|v(t)|\). When \(-t^2 + 1 = 0\), \(t^2 = 1\), this gives \(t = 1\), which is our transition point. We integrate separately for \([0, 1]\) and \([1, 2]\):\[\int_{0}^{1} (t^2 - 1) \, dt = \left[ -\frac{t^3}{3} + t \right]_{0}^{1} = \left(-\frac{1}{3} + 1 \right) = \frac{2}{3}\]\[\int_{1}^{2} (-t^2 + 1) \, dt = \left[ -\frac{t^3}{3} + t \right]_{1}^{2} = \left(-\frac{8}{3} + 2 \right) - \left(-\frac{1}{3} + 1 \right) = \frac{1}{3}\]Total distance is \(\frac{2}{3} + \frac{1}{3} = 1\).

Step 5: Calculate Average Speed

Average speed is calculated by dividing the total distance by the total time:\[\text{Average Speed} = \frac{\text{Total Distance}}{\text{Time Interval}} = \frac{1}{2} = 0.5 \text{ ft/s}\]

Key Concepts

Velocity Functions

A velocity function, like the one in this exercise, describes how fast an object is moving and in which direction at any given time. This function is usually defined in terms of time, represented as \( v(t) \). For example, in the problem, we have the velocity function \( v(t) = -t^2 + 1 \), where \( t \) stands for time in seconds, and the output is feet per second.

• When the value of \( v(t) \) is positive, the object moves forward.
• If \( v(t) \) is negative, the object moves backward.
• A function can be zero, implying the object is momentarily at rest.

Understanding these aspects will help differentiate between when an object is accelerating and slowing down or reversing its direction. It's essential to focus on the sign of the velocity to grasp these changes.

Displacement

Displacement refers to the total change in position of an object over a time interval. It computes how far from the starting point an object ends up, taking into account its direction of travel.
To find displacement using a velocity function, integration is the key. Here, to calculate displacement from \( t = 0 \) to \( t = 2 \), we integrate \( v(t) \) as follows:\[\text{Displacement} = \int_{0}^{2} (-t^2 + 1) \, dt = \left[ -\frac{t^3}{3} + t \right]_{0}^{2} = \frac{-2}{3}\]This result reveals that the object has moved \( \frac{-2}{3} \) feet backward as the final position is less than the initial one (considering the minus sign). Remember, displacement can be negative when direction is reversed.

Average Speed

Average speed is different from average velocity as it considers the entire journey's distance traveled without regard to direction. Here, it involves calculating total distance instead of displacement.
Use the absolute value of the velocity function to find the total distance:- Identify points where velocity changes direction (\( t = 1 \) in this exercise).- Integrate separately for segments where velocity is continuous.Hence, the calculations are:\[\int_{0}^{1} |v(t)| \, dt = \frac{2}{3} , \int_{1}^{2} |v(t)| \, dt = \frac{1}{3}.\]Adding these gives the distance travelled as \( 1 \) foot.
The average speed calculated over the interval is:\[\text{Average Speed} = \frac{1}{2} = 0.5 \text{ ft/s}.\]Average speed is always a positive value since it doesn't regard direction.

Integration in Calculus

Integration is a fundamental calculus operation used to find areas under curves, which is crucial for calculating quantities like displacement. The antiderivative of a function represents the accumulation of quantities, such as distance.

• It involves finding a function whose derivative matches the integrand (function we want to integrate).
• For example, the integration of \( -t^2 + 1 \) results in \( -\frac{t^3}{3} + t \).
• Applying integration over a definite interval finds accumulated values such as total displacement.

Integration helps transform instantaneous velocities into total traveled distances or displacements, providing insight into the motion of objects over time. Understanding this process is key for solving problems related to motion and physical changes over intervals.