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Magazine Chapter 5: Problem 11
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.
Short Answer
Expert verified
The car traveled 250 meters from \(t=0\) to \(t=10\) seconds.
Step by step solution
01
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.
02
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.
03
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: \[ D = \int_{0}^{10} 5t \, dt \]
04
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:\[ D = \frac{500}{2} - 0 = 250 \]
05
Conclusion
The exact distance the car traveled from \( t=0 \) to \( t=10 \) seconds is 250 meters.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.
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:
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: \( \left[ \frac{5(10)^2}{2} \right] - \left[ \frac{5(0)^2}{2} \right] = 250 \).
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:
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.
