Problem 20 - Calculus Early Transcendentals

Question

A cyclist rides down a long straight road at a velocity (in \(\mathrm{m} / \mathrm{min}\) ) given by \(v(t)=400-20 t,\) for \(0 \leq t \leq 10\). a. How far does the cyclist travel in the first 5 min? b. How far does the cyclist travel in the first 10 min? c. How far has the cyclist traveled when her velocity is \(250 \mathrm{m} / \mathrm{min} ?\)

Step-by-Step Solution

Answer

a) In the first 5 minutes: 1000 meters b) In the first 10 minutes: 2000 meters c) When her velocity was 250 m/min: 1312.5 meters

Step 1: Write down the given velocity function

The velocity function is given by \(v(t) = 400 - 20t\), for \(0 \leq t \leq 10\).

Step 2: Integrate the velocity function with respect to time

To find the distance traveled, integrate the velocity function with respect to time to get the displacement function \(s(t)\). $$s(t) = \int (400 - 20t) \ dt$$

Step 3: Evaluate the indefinite integral

Evaluate the indefinite integral to find \(s(t)\): $$s(t) = 400t - 10t^2 + C$$ Where C is the integration constant.

Step 4: Determine the constant of integration

Since the cyclist starts at time \(t=0\), it means their initial displacement is also 0. So, let's set \(t=0\) and \(s(0)=0\) to find the constant C: $$s(0) = 400(0) - 10(0)^2 + C = 0$$ Thus, \(C=0\). Which gives us the displacement function \(s(t) = 400t - 10t^2\).

Step 5: Evaluate the definite integral from t=0 to t=5

Now, find the distance traveled in the first 5 minutes by evaluating the definite integral from 0 to 5: $$s(5) = \int_{0}^{5} (400 - 20t) \ dt = \left[ 400t - 10t^2\right]_{0}^{5}$$ $$s(5) = (400(5) - 10(5)^2) - (400(0) - 10(0)^2) = 1000 \ \text{m}$$ So, the cyclist travels 1000 meters in the first 5 minutes. b. Find the distance traveled by the cyclist in the first 10 minutes

Step 1: Evaluate the definite integral from t=0 to t=10

Use the same displacement function and evaluate from t=0 to t=10: $$s(10) = \int_{0}^{10} (400 - 20t) \ dt = \left[ 400t - 10t^2\right]_{0}^{10}$$ $$s(10) = (400(10) - 10(10)^2) - (400(0) - 10(0)^2) = 2000 \ \text{m}$$ So, the cyclist travels 2000 meters in the first 10 minutes. c. Find the distance traveled by the cyclist when her velocity is 250 m/min

Step 1: Find the time when the velocity is 250 m/min

Use the velocity function \(v(t) = 400 - 20t\) and set \(v(t)=250\) to find the time: $$250 = 400 - 20t$$ $$20t = 150$$ $$t = 7.5 \ \text{minutes}$$

Step 2: Calculate the distance traveled at t=7.5 minutes

Use the displacement function \(s(t) = 400t - 10t^2\) and substitute \(t=7.5\): $$s(7.5) = 400(7.5) - 10(7.5)^2 = 1312.5 \ \text{m}$$ When the cyclist's velocity is 250 m/min, she has traveled 1312.5 meters.

Key Concepts

Velocity Function

Understanding the velocity function is key to analyzing motion problems. In this exercise, the cyclist's velocity function is given as \(v(t) = 400 - 20t\). This function describes how fast the cyclist is traveling at any given time \(t\) within the interval \(0 \leq t \leq 10\). Here, the velocity is expressed in meters per minute.

• When the time \(t = 0\), the velocity is highest at 400 meters per minute.
• As \(t\) increases, the velocity decreases at a steady rate of 20 meters per minute squared.

This type of velocity function, which linearly decreases with time, is common when forces such as resistance slow down motion. It's important to understand that when working with velocity functions, the rate of change of speed over time is determined by the coefficients and terms of the function in question.

Displacement Function

The displacement function describes the total distance traveled by the cyclist over a time interval. To obtain this from the velocity function, we integrate to find \(s(t) = \int (400 - 20t) \, dt\). This becomes \(s(t) = 400t - 10t^2 + C\). Because the cyclist starts at rest with zero displacement, we identify the constant \(C\) as zero, simplifying our displacement function to \(s(t) = 400t - 10t^2\). Using this displacement function allows us to compute the distance covered:

• From \(t = 0\) to \(t = 5\), the displacement is 1000 meters.
• Between \(t = 0\) and \(t = 10\), the full distance is 2000 meters.
• When the cyclist's velocity is 250 meters per minute at \(t = 7.5\), the displacement is 1312.5 meters.

This illustrates how integration transforms velocity functions into displacement functions, offering insights into distances traveled over time.

Cyclist Motion

The cyclist's motion is characterized by a slowing pace over time. Starting off quickly with an initial velocity of 400 meters per minute, the cyclist slows down as time progresses due to a decrease in velocity by 20 units per minute.

• In the first 5 minutes, a significant distance of 1000 meters is covered.
• In the next 5 minutes, the cyclist covers an additional 1000 meters, reaching a total of 2000 meters in 10 minutes.
• At the distinct time when her speed is 250 meters per minute, she has traveled 1312.5 meters, showcasing how velocity directly impacts distance.

Analyzing this motion gives us insight into how external factors, like resistance, impact progress. Cyclists, or any moving object, will often start with a strong velocity which gradually decreases over time due to forces acting against movement.