Problem 11 - Applied Calculus

Question

The velocity of a car is \(f(t)=5 t\) meters/sec. Use a graph of \(f(t)\) to find the exact distance traveled by the car, in meters, from \(t=0\) to \(t=10\) seconds.

Step-by-Step Solution

Answer

The car traveled 250 meters from \(t=0\) to \(t=10\) seconds.

Step 1: Identify the Problem

We need to find the total distance traveled by a car over a given period using its velocity function. Here, the velocity function is given as \(f(t)=5t\), and we need to evaluate the total distance from \(t=0\) to \(t=10\) seconds.

Step 2: Understanding the Function Graphically

The function \( f(t) = 5t \) is a linear function with a slope of 5 and an intercept of 0. This means that the graph is a straight line passing through the origin (0,0). The slope indicates that for every 1 second increase in time, the velocity increases by 5 meters/sec.

Step 3: Integrate to Find Distance

The distance traveled by the car is the area under the velocity curve from \( t=0 \) to \( t=10 \). This can be calculated using the definite integral of the function \( f(t) \) over the interval \([0, 10]\). Thus, the distance \( D \) is given by: \[ D = \int_{0}^{10} 5t \, dt \]

Step 4: Calculate the Integral

Calculate the integral: \[ \int 5t \, dt = \frac{5t^2}{2} + C \] Now, evaluate it from \( t=0 \) to \( t=10 \):\[ D = \left[ \frac{5(10)^2}{2} \right] - \left[ \frac{5(0)^2}{2} \right] \]Calculating this gives:\[ D = \frac{500}{2} - 0 = 250 \]

Step 5: Conclusion

The exact distance the car traveled from \( t=0 \) to \( t=10 \) seconds is 250 meters.

Key Concepts

Velocity Function

In calculus, a velocity function describes how an object's speed changes over time. It typically relates the velocity (speed with direction) to time or another independent variable. For a car whose velocity is described by the function \( f(t) = 5t \), it's important to understand what this function tells us.
The equation \( f(t) = 5t \) means that the car's velocity increases by 5 meters per second every second.
This is a simple linear relationship where the velocity at any time \( t \) can be found by multiplying \( t \) by 5.

• When \( t = 0 \), \( f(0) = 0 \) meters/second, which means the car is at rest.
• When \( t = 1 \), \( f(1) = 5 \) meters/second.
• When \( t = 10 \), \( f(10) = 50 \) meters/second. This indicates that by 10 seconds, the car reaches a speed of 50 meters per second.

Understanding the velocity function helps in predicting future behavior of the moving object and is essential for finding the distance traveled.

Distance Traveled

The distance traveled by an object is generally the total length it covers over a period of time. To determine this distance using a velocity function, we look at the area under the velocity-time graph.
For instance, with the velocity function \( f(t) = 5t \), the graph is a straight line starting at the origin with a slope of 5.
To find the exact distance from \( t=0 \) to \( t=10 \) seconds, calculate the area under this line.

• This area represents the accumulation of velocity, or effectively, how much ground the car covered in the given time span.
• The faster the car goes, the larger the distance traveled in the same unit of time.

Using calculus, specifically integrals, allows us to calculate this area exactly by determining a finite value that represents the total distance.

Definite Integral

In calculus, the definite integral of a function over a specific interval gives the 'accumulated total' or net value of that function within that range.
For the velocity function \( f(t) = 5t \), the integral from \( t=0 \) to \( t=10 \) tells us the total distance the car traveled in that time.
The definite integral is written as \( \int_{0}^{10} 5t \, dt \).
This integral calculates the area under the curve of the velocity function between \( t = 0 \) and \( t = 10 \), which directly translates to distance in this context.
Here's how it works:

• The expression \( \frac{5t^2}{2} \) is derived from integrating \( 5t \).
• Substitute the upper limit (10) and lower limit (0) into this new expression.
• Compute the difference to find the total distance: \( \left[ \frac{5(10)^2}{2} \right] - \left[ \frac{5(0)^2}{2} \right] = 250 \).

Thus, the car travels exactly 250 meters.

Linear Function

A linear function is a type of function that creates a straight line when graphed on the coordinate plane. The general form of a linear function is \( f(t) = mt + b \), where \( m \) is the slope, and \( b \) is the y-intercept.
In our exercise, \( f(t) = 5t \) is a linear function with a slope \( m = 5 \) and intercept \( b = 0 \). This means:

• The function starts at the origin (0,0) and moves upward at a constant rate, illustrating a constant acceleration.
• The slope, 5, tells us how steeply the line rises: for each 1 second increase in time, the velocity increases by 5 meters/second.
• Since there is no intercept other than the origin, the function simplifies calculations regarding area and integration.

The properties of linear functions like these make them easy to work with when modeling real-world phenomena such as motion, where rates of change are constant.