Problem 8 - Basic Technical Mathematics with Calculus

Question

A cyclist goes downhill for 15 min with a velocity to ling \(v=40+64 t(\text { in } \mathrm{km} / \mathrm{h}),\) and then maintains the speed at the bottom for another 30 min. How far does the cyclist go in the 45 min?

Step-by-Step Solution

Answer

The cyclist travels a total of 40 km in 45 minutes.

Step 1: Understanding the Problem

The cyclist travels with a changing velocity over the first 15 minutes and then travels at constant velocity for the next 30 minutes. We need to calculate the total distance covered during these 45 minutes.

Step 2: Convert Time to Hours

Since velocity is in km/h, first convert the time from minutes to hours. We have 15 minutes and 30 minutes. Convert them as follows: \[ t_1 = \frac{15}{60} = 0.25 \text{ hours} \] \[ t_2 = \frac{30}{60} = 0.5 \text{ hours} \]

Step 3: Determine and Integrate Velocity Function

The velocity function given is \( v(t) = 40 + 64t \). To find the distance traveled in the first 15 minutes, integrate this function from 0 to 0.25 hours. The integral is: \[ \int_0^{0.25} (40 + 64t) \, dt \]

Step 4: Calculate Integration

Now, calculate the integral \[ \int_0^{0.25} (40 + 64t) \, dt = \left[ 40t + 32t^2 \right]_0^{0.25} \] Substituting the limits, we get: \[ \left( 40 \times 0.25 + 32 \times (0.25)^2 \right) - \left( 40 \times 0 + 32 \times 0^2 \right) = 10 + 2 = 12 \text{ km} \]

Step 5: Calculate Distance at Constant Speed

After reaching the bottom of the hill, the cyclist maintains the speed attained at 15 minutes for the remaining 30 minutes. Find this speed by evaluating the velocity function at \( t = 0.25 \): \[ v(0.25) = 40 + 64 \times 0.25 = 56 \text{ km/h} \] Now calculate the distance traveled in the next 0.5 hours: \[ \text{Distance} = v \times t_2 = 56 \times 0.5 = 28 \text{ km} \]

Step 6: Add the Distances

Add the distance covered during the acceleration phase and the constant speed phase. \[ 12 + 28 = 40 \text{ km} \]

Key Concepts

Understanding Integration

Integration is a fundamental concept in calculus used to calculate various quantities, such as areas under curves, accumulated values, or total distances. When you integrate a velocity function, you determine the total distance traveled over a period of time. The process involves finding the antiderivative of the velocity function and evaluating it over the given interval.

• Integration accumulates small, instantaneous changes over a span of time, like calculating total distance from varying speeds.
• The notation for integration includes the integral sign \( \int \), the function to integrate, and the differential \( dt \) which indicates small changes in time.
• Apply the limits of integration, here for instance, from 0 to 0.25, to find the total change over that interval.

For example, integrating the function \( v(t) = 40 + 64t \) from 0 to 0.25 hours gives us the distance traveled during the initial phase of acceleration.

Velocity Function Explained

The velocity function describes how an object's speed changes over time. In this problem, the cyclist's velocity is given by the formula \( v(t) = 40 + 64t \), where \( v \) is in km/h and \( t \) is in hours.

• Here, the constant 40 represents the initial velocity when time \( t \) is zero.
• The term 64t indicates that the cyclist is accelerating; the velocity increases by 64 km/h every hour.
• When integrating this function, it helps to determine the change in speed over a specific time frame.

At \( t = 0.25 \) hours, substitute into the velocity function to find that the speed is 56 km/h, which determines the speed maintained in the next time segment.

Calculating Distance Using Integration and Constant Speed

To calculate distance using integration, apply the velocity function over the specific time frames. First, integrate the velocity function for the acceleration period: - This calculates the area under the velocity-time graph, giving you the distance traveled during the first phase.- In this example, \[ \int_0^{0.25} (40 + 64t) \, dt \] results in a distance of 12 km.Next, during the constant speed phase, calculate the distance as the product of the velocity and time:- The cyclist's speed at \( t = 0.25 \) hours is 56 km/h. - Maintaining this speed for 0.5 hours, compute the distance as \[ \text{Distance} = 56 \times 0.5 = 28 \text{ km} \].Combine these distances to find the total: - During the viral velocity change, the cyclist traveled 12 km. - During constant speed, covered an additional 28 km.Thus, the total distance traveled is 40 km.