Problem 42 - Calculus and its applications

Question

Suppose that in \(t\) hours, a truck travels \(s(t)\) miles, where $$ s(t)=10 t^{2} $$ a) Find \(s(5)-s(2)\). What does this represent? b) Find the average rate of change of distance with respect to time as \(t\) changes from \(t_{1}=2\) to \(t_{2}=5\). This is also average velocity.

Step-by-Step Solution

Answer

The truck travels 210 miles between 2 and 5 hours with an average velocity of 70 mph.

Step 1: Calculate Distance at t=5

Substitute 5 into the function to find the distance at that time: \[ s(5) = 10 imes 5^2 = 10 imes 25 = 250 \] So, the distance traveled at \( t = 5 \) hours is 250 miles.

Step 2: Calculate Distance at t=2

Substitute 2 into the function to find the distance at that time: \[ s(2) = 10 imes 2^2 = 10 imes 4 = 40 \] So, the distance traveled at \( t = 2 \) hours is 40 miles.

Step 3: Find the Change in Distance

Calculate the difference in distance by subtracting distances at \( t = 2 \) and \( t = 5 \): \[ s(5) - s(2) = 250 - 40 = 210 \] This represents the distance traveled by the truck between 2 and 5 hours, which is 210 miles.

Step 4: Calculate the Average Rate of Change

Use the formula for average rate of change (or average velocity) over the interval from \( t_1 = 2 \) to \( t_2 = 5 \): \[ \text{Average Rate of Change} = \frac{s(5) - s(2)}{5 - 2} = \frac{210}{3} = 70 \] This means the average velocity of the truck between 2 and 5 hours is 70 miles per hour.

Key Concepts

Distance Function

The distance function is a way to understand how far an object travels over time. In our scenario, the truck's distance over time is given by the quadratic function \( s(t) = 10t^2 \). This essentially models the distance in miles the truck travels in \( t \) hours. To make sense of the function, notice that it calculates where the truck is at any given time \( t \). The function \( s(t) \) embodies the relationship between time and distance, simplifying the process of predicting future positions on a route the truck takes. For example, if you plug in \( t = 5 \) to find \( s(5) \), it shows the truck's position after 5 hours. Thus, understanding this function helps determine how far the truck has traveled over a specified period. By knowing how to use the distance function, one can efficiently track or plan journeys based on time and pace.

Average Velocity

Average velocity is a concept that helps you understand the overall speed of an object throughout a journey. It's calculated using the average rate of change formula. This formula looks at how much distance has changed over a specific period. By evaluating velocity as an average, you can capture a general sense of speed over that interval. For the truck traveling from \( t_1 = 2 \) to \( t_2 = 5 \), we use:

• Change in distance: \( s(5) - s(2) \), which is 210 miles.
• Duration: \( 5 - 2 \), which gives 3 hours.

Calculating average velocity involves dividing the distance by the time, resulting in \( 70 \) miles per hour. Therefore, the truck's average velocity captures how fast, on average, it travels between those two points in time.

Quadratic Functions

Quadratic functions are mathematical expressions that describe a parabolic curve. Commonly written in the form \( ax^2 + bx + c \), they can effectively model various real-life scenarios, such as the path of a projectile or, in this case, the distance covered by a vehicle over time. The truck's distance function \( s(t) = 10t^2 \) is a simple quadratic function. Here:

• \( a = 10 \)
• \( b = 0 \)
• \( c = 0 \)

These functions often indicate changing rates of speed and are crucial in predicting positions or velocities at certain points in time. A quadratic function's analysis reveals how much the response variable, or in our case, the distance, changes proportionately with time squared. Grasping quadratic functions enriches understanding of dynamic movements like acceleration and can be often applied beyond just determining distances.