Velocity Function of an Automobile: Maximum Speed, Zero Velocity, and Distance Traveled

Math Problem Statement

The velocity of a (fast) automobile on a straight highway is given by the function to the right, where t is measured in seconds and v has units of m/s.

v left parenthesis t right parenthesis equals left brace Start 3 By 2 Matrix 1st Row 1st Column 3 t 2nd Column if 0 less than or equals t less than 20 2nd Row 1st Column 60 2nd Column if 20 less than or equals t less than 45 3rd Row 1st Column 240 minus 4 t 2nd Column t greater than or equals 45 EndMatrixv(t)=

if 0≤t<20

if 20≤t<45

240−4t

t≥45

a. Graph the velocity function for

0less than or equals≤tless than or equals≤100100.

When is the velocity a maximum? When is the velocity zero?

b. What is the distance traveled by the automobile in the first

3535

c. What is the distance traveled by the automobile in the first

6060

d. What is the position of the automobile when

tequals=7575?

Solution

The velocity function $v(t)$ of the automobile is defined piecewise as:

\begin{cases} 3t & \text{if } 0 \leq t < 20 \\ 60 & \text{if } 20 \leq t < 45 \\ 240 - 4t & \text{if } t \geq 45 \end{cases}$$ We will now go through each part of the problem step by step. ### **Part (a): Graph the velocity function** To graph the velocity function for $$0 \leq t \leq 100$$, we can break it into three regions: 1. $$v(t) = 3t$$ for $$0 \leq t < 20$$: This is a linear function with slope 3. 2. $$v(t) = 60$$ for $$20 \leq t < 45$$: This is a constant function at 60. 3. $$v(t) = 240 - 4t$$ for $$t \geq 45$$: This is a linear function with negative slope (-4). We can graph these functions accordingly. #### **Maximum velocity** The velocity reaches its maximum when $$v(t) = 60$$ for $$20 \leq t < 45$$. The constant function 60 is the highest point in the graph. #### **Zero velocity** To find when the velocity is zero, we solve for $$v(t) = 0$$. This can only happen in the third region where $$v(t) = 240 - 4t$$: $$240 - 4t = 0 \quad \Rightarrow \quad t = 60$$ So, the velocity is zero at $$t = 60 \, \text{s}$$. ### **Part (b): Distance traveled in the first 35 seconds** The distance traveled is the integral of the velocity function. To find the distance traveled in the first 35 seconds, we break it into two intervals: - From $$0 \leq t < 20$$, the velocity is $$3t$$. - From $$20 \leq t \leq 35$$, the velocity is constant at 60. The distance traveled is: $$d = \int_0^{20} 3t \, dt + \int_{20}^{35} 60 \, dt$$ We can calculate this. ### **Part (c): Distance traveled in the first 60 seconds** Again, we need to integrate the velocity function from $$t = 0$$ to $$t = 60$$. We break it into three intervals: - From $$0 \leq t < 20$$, $$v(t) = 3t$$. - From $$20 \leq t < 45$$, $$v(t) = 60$$. - From $$45 \leq t \leq 60$$, $$v(t) = 240 - 4t$$. The total distance is: $$d = \int_0^{20} 3t \, dt + \int_{20}^{45} 60 \, dt + \int_{45}^{60} (240 - 4t) \, dt$$ We can calculate this as well. ### **Part (d): Position of the automobile when $$t = 75$$** The position of the automobile at $$t = 75$$ is the total distance traveled from $$t = 0$$ to $$t = 75$$. This requires integrating the velocity over three intervals: $$d = \int_0^{20} 3t \, dt + \int_{20}^{45} 60 \, dt + \int_{45}^{75} (240 - 4t) \, dt$$ Now, I'll calculate the integrals and provide you with the distances for parts (b), (c), and (d). Let's compute these values.Here are the results for the distance traveled: - **Part (b):** The distance traveled in the first 35 seconds is **1500 meters**. - **Part (c):** The distance traveled in the first 60 seconds is **2550 meters**. - **Part (d):** The position of the automobile at $$t = 75$$ seconds is **2100 meters**. Would you like any further details or clarifications? Here are some related questions: 1. How can we generalize the formula for distance using velocity as a function of time? 2. How does changing the initial or final velocities affect the total distance? 3. How would the graph of velocity vs. time change if the car decelerated sooner? 4. What happens if the velocity reaches zero multiple times in a different scenario? 5. How would we calculate average velocity over any arbitrary interval? **Tip:** The area under the velocity-time graph represents the distance traveled.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Piecewise functions
Velocity-time graph
Integration
Distance as an integral of velocity

Formulas

v(t) = { 3t if 0 ≤ t < 20, 60 if 20 ≤ t < 45, 240 - 4t if t ≥ 45 }
Distance = ∫ v(t) dt

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 (Calculus and Advanced Physics)

Related Recommendation

AP Calculus Kinematics and Velocity-Graph Problems

Kinematics: Analyzing Velocity and Acceleration of a Car

Interpreting Graphs of Piecewise Functions: Finding Function Values and Rates of Change

Vehicle Motion: Velocity and Position from Graphs

Solving a Piecewise Function: Sketch, Range, and Discontinuity