Question

The velocity of a (fast) automobile on a straight highway is given by the function $$v(t)=\left\\{\begin{array}{ll}3 t & \text { if } 0 \leq t<20 \\\60 & \text { if } 20 \leq t<45 \\\240-4 t & \text { if } t \geq 45,\end{array}\right.$$ where \(t\) is measured in seconds and \(v\) has units of meters/second. a. Graph the velocity function, for \(0 \leq t \leq 70 .\) When is the velocity a maximum? When is the velocity zero? b. What is the distance traveled by the automobile in the first 30 s? c. What is the distance traveled by the automobile in the first \(60 \mathrm{s} ?\) d. What is the position of the automobile when \(t=75 ?\)

Short answer

Also, find the distance traveled during the first 30 seconds, the first 60 seconds, and the position of the automobile when t = 75 seconds. Answer: The maximum velocity of the automobile occurs during the interval 20 ≤ t < 45 seconds, with a maximum speed of 60 meters/second. The velocity reaches zero at t = 60 seconds. The distance traveled during the first 30 seconds is 1200 meters, during the first 60 seconds is 2550 meters, and the position of the automobile at t = 75 seconds is 2925 meters from its initial location.

Step-by-step solution

Graph the velocity function

To graph the function, we need to consider all three cases defined in the piecewise function separately: 1. For \(0 \leq t < 20\), the function is \(v(t) = 3t\). 2. For \(20 \leq t < 45\), the function is a constant \(v(t) = 60\). 3. For \(t \geq 45\), the function is \(v(t) = 240 - 4t\). Plot these functions individually on the same graph and combine them to get the complete graph of the velocity function for \(0 \leq t \leq 70\).

Determine when the velocity is maximum

Looking at the graph created in Step 1, we can see that the maximum occurs during the constant phase when \(20 \leq t < 45\). During this duration, the maximum velocity is 60 meters/second.

Determine when the velocity is zero

The velocity is zero when the function \(v(t)\) is zero. This can only happen in the third case, where \(v(t) = 240 - 4t\). Set the function to zero and solve for t: $$240 - 4t = 0 \Rightarrow t = 60$$ #b. Distance traveled during the first 30 seconds#

Find distance traveled during each interval

We need to calculate the distance traveled since the beginning and until t = 30 seconds. This duration covers two intervals of the piecewise function: 1. For \(0 \le t < 20\): Integrate the function \(v(t) = 3t\) from 0 to 20: $$\int_0^{20} 3t dt$$ 2. For \(20 \leq t < 30\): Integrate the function \(v(t) = 60\) from 20 to 30: $$\int_{20}^{30} 60 dt$$

Add up the distances from both intervals

The total distance traveled during the first 30 seconds is the sum of distances in both intervals: $$\int_0^{30} v(t) dt = \int_0^{20} 3t dt + \int_{20}^{30} 60 dt$$

Evaluate the integrals and find the distance

Evaluate the integrals obtained in Step 2 for each interval and find the total distance traveled: $$\int_0^{20} 3t dt = \frac{3}{2}t^2\Big|_0^{20} = 600$$ $$\int_{20}^{30} 60 dt = 60t\Big|_{20}^{30} = 600$$ The total distance traveled during the first 30 seconds is: $$600 + 600 = 1200 \mathrm{meters}$$ #c. Distance traveled during the first 60 seconds#

Define the intervals

Same as before, to find the distance traveled we need to integrate the appropriate pieces of the function and add them all together. In this case, we need to cover all 60 seconds intervals: 1. \(0 \leq t < 20\) 2. \(20 \leq t < 45\) 3. \(45 \leq t < 60\)

Integrate all intervals individually and find the distance

Integrate the function for each interval and add the distances together. $$\int_0^{60} v(t) dt = \int_0^{20} 3t dt + \int_{20}^{45} 60 dt + \int_{45}^{60} (240 - 4t) dt $$ Evaluate them individually: 1. We already found the integral for the first interval in part b: 600 meters 2. Second interval: $$\int_{20}^{45} 60 dt = 60t\Big|_{20}^{45} = 1500$$ 3. Third interval: $$\int_{45}^{60} (240-4t) dt = (240t - 2t^2)\Big|_{45}^{60} = 450$$ Now, sum the distances for all three intervals: $$600 + 1500 + 450 = 2550 \mathrm{meters}$$ The total distance traveled during the first 60 seconds is 2550 meters. #d. Position of the automobile when t = 75 seconds#

Calculate and sum distances traveled from all intervals until t=75

Similar to the previous parts, we should consider all intervals until we reach \(t=75\) seconds. For the first 60 seconds, we already found the distance traveled to be 2550 meters. Find the distance traveled during the last interval \(60 \leq t < 75\): $$\int_{60}^{75} (240 - 4t) dt $$

Evaluate the integral and sum distances

Evaluate the integral for the last interval: $$\int_{60}^{75} (240-4t) dt = (240t - 2t^2)\Big|_{60}^{75} = 375$$ Now add the distances traveled in 60 seconds and from 60 seconds to 75 seconds: $$2550 + 375 = 2925 \mathrm{meters}$$ The position of the automobile when \(t = 75\) seconds is 2925 meters from its initial location.

Key concepts

Understanding Piecewise Functions

Piecewise functions are an essential concept in calculus that help describe different behaviors of a function over multiple intervals. These functions are defined by different expressions for various domains.
In the context of velocity, a piecewise function can model different speeds at different times. For example, a car might accelerate over a period, move at a constant speed, then decelerate.

• The piecewise function for the car's velocity was \(v(t)=\left\{\begin{array}{ll}3 t & \text { if } 0 \leq t<20 \60 & \text { if } 20 \leq t<45 \240-4 t & \text { if } t \geq 45,\end{array}\right.\), capturing the car's acceleration, cruising, and deceleration phases.
• By breaking down velocity into intervals based on time, we can analyze each segment independently, thus simplifying complex motions into understandable parts.
• In this way, piecewise functions highlight the importance of evaluating a function's behavior at different stages, providing insights into the real-world movement of objects.

How to Calculate Distance Traveled Using Velocity

Distance traveled can be found by determining the area under the velocity-time curve. This is critical in physics and engineering as it bridges the concept of velocity and physical displacement.
To calculate the total distance traveled within a certain period, integrate the velocity function over that time period.

• Integrals of piecewise velocity functions need to be performed over the individual intervals, i.e., calculate the area under each segment separately.
• For the first 30 seconds, the car travels two segments: accelerating from \(0\) to \(20\) seconds with \(v(t)=3t\) and cruising from \(20\) to \(30\) seconds with \(v(t)=60\).
• Evaluate and sum these integrals to find the total distance traveled during each interval.
  • The integral gives the area under each segment of the curve, translating the motion information into distance.

The integration of these intervals helps in visualizing the changes in motion and quantifying distances over time.

Definite Integrals in Finding Total Distance

A definite integral is a fundamental tool in calculus, mainly used to compute the accumulation of quantities over a defined interval. In terms of velocity functions, definite integrals help find the total distance traveled over time.
To evaluate definite integrals:

• Take the integral of each individual section of a piecewise function over its interval.
• Combine these results to find the total value over a specified period.

For example:

• When finding the distance traveled during the first 60 seconds, sum up the definite integrals of each section:
  • \(\int_0^{20} 3t \, dt\), \(\int_{20}^{45} 60 \, dt\), and \(\int_{45}^{60} (240-4t) \, dt\).

Each definite integral gives the distance for its interval, and summing them provides the total distance for the entire period.
This process of calculation helps students understand how definite integrals function in practical applications, reinforcing the concept's usefulness beyond theoretical exercises.

Finding Maximum and Minimum Values of a Function

Determining the maximum and minimum values of a function is crucial in analyzing the behavior of any given process.
For velocity pieces functions:

• A maximum value identifies the highest velocity attained during the movement.
• A minimum identifies stationary or no movement when velocity is zero or the lowest.

In the given example:

• The velocity reaches its maximum of \(60\) meters per second during the interval \(20 \leq t < 45\), shown as a constant segment on the graph.
• The velocity is zero at \(t = 60\) when the function \(v(t) = 240 - 4t\) results in zero. \(240 - 4(60) = 0\)

Evaluating these extremities in a piecewise velocity function helps pinpoint critical points in motion analysis. Understanding these concepts guarantees a clear view of transient phases in dynamic systems.