Problem 4
Question
Explain in words what the integral represents and give units. \(\int_{0}^{6} a(t) d t\), where \(a(t)\) is acceleration in \(\mathrm{km} / \mathrm{hr}^{2}\) and \(t\) is time in hours.
Step-by-Step Solution
Verified Answer
The integral represents the change in velocity over 6 hours, with units of \( \mathrm{km/hr} \).
1Step 1: Understand the Integral
The integral \( \int_{0}^{6} a(t) \, dt \) represents the area under the curve of the function \( a(t) \) from \( t = 0 \) to \( t = 6 \). This essentially calculates the total change in velocity over this time period.
2Step 2: Analyze the Function
The function \( a(t) \) is given in units of \( \mathrm{km/hr^2} \), which means that it describes how the velocity of an object changes with respect to time (acceleration).
3Step 3: Determine the Units
To find the units of the integral, consider the dimensions of the integrand and the differential part, \( dt \). The function \( a(t) \, dt \) has units \( \mathrm{km/hr^2} \times \mathrm{hours} = \mathrm{km/hr} \). This indicates that the integral represents a change in velocity.
4Step 4: Conclude the Meaning
Thus, the integral \( \int_{0}^{6} a(t) \, dt \) calculates the change in velocity from the start time \( t = 0 \) to the end time \( t = 6 \ hours \). The resultant value is in \( \mathrm{km/hr} \), indicative of a velocity.
Key Concepts
AccelerationChange in VelocityUnits of Measurement
Acceleration
Acceleration describes how the velocity of an object changes over time. When we talk about acceleration in integral calculus, it's often represented by the function \( a(t) \), where \( t \) stands for time. Acceleration measures how quickly or slowly an object's speed is increasing or decreasing.
Think of acceleration as a car's speedometer needle moving. If the needle goes up quickly, that's a high acceleration. If it's slow, the acceleration is low. This is crucial in physics because it helps us understand the dynamics of motion.
In equations, acceleration is typically expressed in units like \( \text{km/hr}^2 \) or \( \text{m/s}^2 \). These units reflect a change in velocity over time. In our integral problem, \( a(t) \) is given in \( \text{km/hr}^2 \).
Think of acceleration as a car's speedometer needle moving. If the needle goes up quickly, that's a high acceleration. If it's slow, the acceleration is low. This is crucial in physics because it helps us understand the dynamics of motion.
In equations, acceleration is typically expressed in units like \( \text{km/hr}^2 \) or \( \text{m/s}^2 \). These units reflect a change in velocity over time. In our integral problem, \( a(t) \) is given in \( \text{km/hr}^2 \).
- Acceleration helps determine the change in velocity over specific time intervals.
- Constant acceleration means a steady change in speed, while variable acceleration implies the rate of change is fluctuating.
Change in Velocity
The concept of change in velocity indicates how the speed of an object shifts over a particular duration.
In calculus, this is often determined by integrating the acceleration function. For example, the integral \( \int_{0}^{6} a(t) \, dt \) calculates the total change in velocity from time \( t = 0 \) to \( t = 6 \).
Let's break that down. By integrating acceleration over time, you accumulate the total change in speed. It's like adding up all the little speed bumps over the course of 6 hours.
Integrating is a powerful tool as it can provide an overall picture of how much the velocity has shifted, not just a snapshot at a particular moment. It gives us the net change.
In calculus, this is often determined by integrating the acceleration function. For example, the integral \( \int_{0}^{6} a(t) \, dt \) calculates the total change in velocity from time \( t = 0 \) to \( t = 6 \).
Let's break that down. By integrating acceleration over time, you accumulate the total change in speed. It's like adding up all the little speed bumps over the course of 6 hours.
Integrating is a powerful tool as it can provide an overall picture of how much the velocity has shifted, not just a snapshot at a particular moment. It gives us the net change.
- The integration process sums up tiny changes over small time intervals to find the total change.
- This concept helps in real-world scenarios like determining how fast a vehicle was traveling after accelerating for a certain period.
Units of Measurement
Units of measurement in physics and calculus are like a language that describes how quantities relate to each other.
In our integral, understanding the units helps specify exactly what the result means in the context of real-world motion. With \( a(t) \) in \( \text{km/hr}^2 \) and \( t \) in hours, the integral \( \int_{0}^{6} a(t) \, dt \) has units \( \text{km/hr}^2 \times \text{hr} = \text{km/hr} \).
This is crucial because it tells us the result of the integral is a velocity change. Knowing the units allows you to interpret the numeric results correctly.
Units are not just labels; they are part of the equation itself and must be consistent throughout calculations. As an analogy, think of units as the currency in math trade. They define what you get after the calculation.
In our integral, understanding the units helps specify exactly what the result means in the context of real-world motion. With \( a(t) \) in \( \text{km/hr}^2 \) and \( t \) in hours, the integral \( \int_{0}^{6} a(t) \, dt \) has units \( \text{km/hr}^2 \times \text{hr} = \text{km/hr} \).
This is crucial because it tells us the result of the integral is a velocity change. Knowing the units allows you to interpret the numeric results correctly.
Units are not just labels; they are part of the equation itself and must be consistent throughout calculations. As an analogy, think of units as the currency in math trade. They define what you get after the calculation.
- Units make sure you mix quantities of the same kind.
- Correct unit conversion guarantees effective communication of scientific results.
Other exercises in this chapter
Problem 3
Find the area between \(y=x+5\) and \(y=2 x+1\) between \(x=0\) and \(x=2\).
View solution Problem 4
Use the following table to estimate \(\int_{3}^{4} W(t) d t\). What are \(n\) and \(\Delta t ?\) $$ \begin{array}{c|c|c|c|c|c|c} \hline t & 3.0 & 3.2 & 3.4 & 3.
View solution Problem 4
Find the area enclosed by \(y-3 x\) and \(y=x^{2}\).
View solution Problem 5
Use the following table to estimate \(\int_{0}^{15} f(x) d x\). $$ \begin{array}{c|c|c|c|c|c|c} \hline x & 0 & 3 & 6 & 9 & 12 & 15 \\ \hline f(x) & 50 & 48 & 44
View solution