A. Find The Average Velocity On The Interval { T=1$}$ To { T=8$}$ Seconds. Give Correct Units.${ \frac{1}{7} \int_1^8 \text{function Here} = 26.42 \text{ (units)} }$B. On What Interval(s) Of Time Is The Particle Moving To

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Introduction

Average velocity is a fundamental concept in physics that describes the rate of change of an object's position with respect to time. It is a crucial quantity in understanding the motion of objects and is used extensively in various fields, including mechanics, thermodynamics, and electromagnetism. In this article, we will explore the concept of average velocity and provide a step-by-step solution to find the average velocity on the interval [t=1] to [t=8] seconds.

What is Average Velocity?

Average velocity is defined as the total displacement of an object divided by the total time taken. Mathematically, it can be represented as:

AverageΒ Velocity=Ξ”xΞ”t{ \text{Average Velocity} = \frac{\Delta x}{\Delta t} }

where Ξ”x is the total displacement and Ξ”t is the total time taken.

Finding the Average Velocity on the Interval [t=1] to [t=8] Seconds

To find the average velocity on the interval [t=1] to [t=8] seconds, we need to know the position function of the object. Let's assume that the position function is given by:

x(t)=2t2+3tβˆ’4{ x(t) = 2t^2 + 3t - 4 }

We are given that the average velocity on the interval [t=1] to [t=8] seconds is equal to 26.42 units. We can use the formula for average velocity to set up an equation:

17∫18x(t)dt=26.42 (units){ \frac{1}{7} \int_1^8 x(t) dt = 26.42 \text{ (units)} }

Evaluating the Integral

To evaluate the integral, we need to find the antiderivative of the position function x(t). Using the power rule of integration, we get:

∫x(t)dt=∫(2t2+3tβˆ’4)dt{ \int x(t) dt = \int (2t^2 + 3t - 4) dt }

=23t3+32t2βˆ’4t+C{ = \frac{2}{3}t^3 + \frac{3}{2}t^2 - 4t + C }

where C is the constant of integration.

Applying the Fundamental Theorem of Calculus

The fundamental theorem of calculus states that the definite integral of a function can be evaluated as the difference between the antiderivative of the function evaluated at the upper limit and the antiderivative evaluated at the lower limit. In this case, we have:

∫18x(t)dt=[23t3+32t2βˆ’4t]18{ \int_1^8 x(t) dt = \left[ \frac{2}{3}t^3 + \frac{3}{2}t^2 - 4t \right]_1^8 }

=(23(8)3+32(8)2βˆ’4(8))βˆ’(23(1)3+32(1)2βˆ’4(1)){ = \left( \frac{2}{3}(8)^3 + \frac{3}{2}(8)^2 - 4(8) \right) - \left( \frac{2}{3}(1)^3 + \frac{3}{2}(1)^2 - 4(1) \right) }

Simplifying the Expression

Simplifying the expression, we get:

=(23(512)+32(64)βˆ’32)βˆ’(23+32βˆ’4){ = \left( \frac{2}{3}(512) + \frac{3}{2}(64) - 32 \right) - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

=(10243+96βˆ’32)βˆ’(23+32βˆ’4){ = \left( \frac{1024}{3} + 96 - 32 \right) - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

=(10243+64)βˆ’(23+32βˆ’4){ = \left( \frac{1024}{3} + 64 \right) - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

Evaluating the Expression

Evaluating the expression, we get:

=10243+64βˆ’23βˆ’32+4{ = \frac{1024}{3} + 64 - \frac{2}{3} - \frac{3}{2} + 4 }

=10243+64βˆ’23βˆ’96+4{ = \frac{1024}{3} + 64 - \frac{2}{3} - \frac{9}{6} + 4 }

=10243+64βˆ’23βˆ’32+4{ = \frac{1024}{3} + 64 - \frac{2}{3} - \frac{3}{2} + 4 }

Simplifying the Expression

Simplifying the expression, we get:

=10243+64βˆ’23βˆ’32+4{ = \frac{1024}{3} + 64 - \frac{2}{3} - \frac{3}{2} + 4 }

=10243+64βˆ’23βˆ’96+4{ = \frac{1024}{3} + 64 - \frac{2}{3} - \frac{9}{6} + 4 }

=10243+64βˆ’46βˆ’96+4{ = \frac{1024}{3} + 64 - \frac{4}{6} - \frac{9}{6} + 4 }

Evaluating the Expression

Evaluating the expression, we get:

=10243+64βˆ’136+4{ = \frac{1024}{3} + 64 - \frac{13}{6} + 4 }

=10243+68βˆ’136{ = \frac{1024}{3} + 68 - \frac{13}{6} }

Simplifying the Expression

Simplifying the expression, we get:

=10243+4086βˆ’136{ = \frac{1024}{3} + \frac{408}{6} - \frac{13}{6} }

=10243+3956{ = \frac{1024}{3} + \frac{395}{6} }

Evaluating the Expression

Evaluating the expression, we get:

=61446+3956{ = \frac{6144}{6} + \frac{395}{6} }

=65396{ = \frac{6539}{6} }

Finding the Average Velocity

Now that we have evaluated the integral, we can find the average velocity on the interval [t=1] to [t=8] seconds:

Average Velocity=17∫18x(t)dt{ \text{Average Velocity} = \frac{1}{7} \int_1^8 x(t) dt }

=17(65396){ = \frac{1}{7} \left( \frac{6539}{6} \right) }

=653942{ = \frac{6539}{42} }

=156.07Β (units){ = 156.07 \text{ (units)} }

Conclusion

In this article, we have found the average velocity on the interval [t=1] to [t=8] seconds. We have used the formula for average velocity and evaluated the integral of the position function x(t) to find the average velocity. The average velocity on the interval [t=1] to [t=8] seconds is equal to 156.07 units.

Introduction

To determine on what interval(s) of time the particle is moving to the right, we need to find the intervals where the velocity function is positive.

What is Velocity?

Velocity is the rate of change of an object's position with respect to time. It is a vector quantity and has both magnitude and direction.

Finding the Velocity Function

To find the velocity function, we need to take the derivative of the position function x(t):

v(t)=dxdt{ v(t) = \frac{dx}{dt} }

=ddt(2t2+3tβˆ’4){ = \frac{d}{dt} (2t^2 + 3t - 4) }

=4t+3{ = 4t + 3 }

Finding the Intervals Where the Velocity Function is Positive

To find the intervals where the velocity function is positive, we need to set the velocity function equal to zero and solve for t:

4t+3=0{ 4t + 3 = 0 }

4t=βˆ’3{ 4t = -3 }

t=βˆ’34{ t = -\frac{3}{4} }

Conclusion

The particle is moving to the right on the interval [t=1, ∞).

Introduction

In this article, we have found the average velocity on the interval [t=1] to [t=8] seconds and determined on what interval(s) of time the particle is moving to the right.

What is the Significance of Average Velocity?

Average velocity is a fundamental concept in physics that describes the rate of change of an object's position with respect to time. It is a crucial quantity in understanding the motion of objects and is used extensively in various fields, including mechanics, thermodynamics, and electromagnetism.

What is the Significance of Velocity?

Velocity is the rate of change of an object's position with respect to time. It is a vector quantity and has both magnitude and direction. The velocity function is used to determine on what interval(s) of time the particle is moving to the right.

Conclusion

In this article, we have found the average velocity on the interval [t=1] to [t=8] seconds and determined on what interval(s) of time the particle is moving to the right. The average velocity on the interval [t=1] to [t=8] seconds is equal to 156.07 units. The particle is moving to the right on the interval [t=1, ∞).

Introduction

In our previous article, we discussed the concept of average velocity and velocity, and how they are used to describe the motion of objects. In this article, we will answer some frequently asked questions about average velocity and velocity.

Q: What is the difference between average velocity and velocity?

A: Average velocity is the total displacement of an object divided by the total time taken, while velocity is the rate of change of an object's position with respect to time.

Q: How do I calculate the average velocity of an object?

A: To calculate the average velocity of an object, you need to know the position function of the object and the time interval over which you want to find the average velocity. You can use the formula:

AverageΒ Velocity=1Ξ”t∫t1t2x(t)dt{ \text{Average Velocity} = \frac{1}{\Delta t} \int_{t_1}^{t_2} x(t) dt }

where x(t) is the position function, Ξ”t is the time interval, and t1 and t2 are the initial and final times.

Q: What is the significance of average velocity?

A: Average velocity is a fundamental concept in physics that describes the rate of change of an object's position with respect to time. It is a crucial quantity in understanding the motion of objects and is used extensively in various fields, including mechanics, thermodynamics, and electromagnetism.

Q: How do I determine on what interval(s) of time the particle is moving to the right?

A: To determine on what interval(s) of time the particle is moving to the right, you need to find the intervals where the velocity function is positive. You can do this by setting the velocity function equal to zero and solving for t.

Q: What is the relationship between average velocity and velocity?

A: The average velocity of an object is equal to the velocity of the object at the midpoint of the time interval.

Q: How do I calculate the velocity function?

A: To calculate the velocity function, you need to take the derivative of the position function x(t):

v(t)=dxdt{ v(t) = \frac{dx}{dt} }

Q: What is the significance of velocity?

A: Velocity is the rate of change of an object's position with respect to time. It is a vector quantity and has both magnitude and direction. The velocity function is used to determine on what interval(s) of time the particle is moving to the right.

Q: Can you give an example of how to calculate the average velocity of an object?

A: Yes, let's say we have an object that moves according to the position function x(t) = 2t^2 + 3t - 4. We want to find the average velocity of the object over the time interval [t=1, t=8]. We can use the formula:

AverageΒ Velocity=1Ξ”t∫t1t2x(t)dt{ \text{Average Velocity} = \frac{1}{\Delta t} \int_{t_1}^{t_2} x(t) dt }

=17∫18(2t2+3tβˆ’4)dt{ = \frac{1}{7} \int_1^8 (2t^2 + 3t - 4) dt }

=17[23t3+32t2βˆ’4t]18{ = \frac{1}{7} \left[ \frac{2}{3}t^3 + \frac{3}{2}t^2 - 4t \right]_1^8 }

=17(23(8)3+32(8)2βˆ’4(8))βˆ’(23(1)3+32(1)2βˆ’4(1)){ = \frac{1}{7} \left( \frac{2}{3}(8)^3 + \frac{3}{2}(8)^2 - 4(8) \right) - \left( \frac{2}{3}(1)^3 + \frac{3}{2}(1)^2 - 4(1) \right) }

=17(10243+64βˆ’32)βˆ’(23+32βˆ’4){ = \frac{1}{7} \left( \frac{1024}{3} + 64 - 32 \right) - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

=17(10243+32)βˆ’(23+32βˆ’4){ = \frac{1}{7} \left( \frac{1024}{3} + 32 \right) - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

=17(10243+963)βˆ’(23+32βˆ’4){ = \frac{1}{7} \left( \frac{1024}{3} + \frac{96}{3} \right) - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

=17(11203)βˆ’(23+32βˆ’4){ = \frac{1}{7} \left( \frac{1120}{3} \right) - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

=112021βˆ’(23+32βˆ’4){ = \frac{1120}{21} - \left( \frac{2}{3} + \frac{3}{2} - 4 \right) }

=112021βˆ’(46+96βˆ’4){ = \frac{1120}{21} - \left( \frac{4}{6} + \frac{9}{6} - 4 \right) }

=112021βˆ’(136βˆ’4){ = \frac{1120}{21} - \left( \frac{13}{6} - 4 \right) }

=112021βˆ’(136βˆ’246){ = \frac{1120}{21} - \left( \frac{13}{6} - \frac{24}{6} \right) }

=112021βˆ’(βˆ’116){ = \frac{1120}{21} - \left( \frac{-11}{6} \right) }

=112021+116{ = \frac{1120}{21} + \frac{11}{6} }

=112021+7742{ = \frac{1120}{21} + \frac{77}{42} }

=224042+7742{ = \frac{2240}{42} + \frac{77}{42} }

=231742{ = \frac{2317}{42} }

=55.17Β (units){ = 55.17 \text{ (units)} }

Conclusion

In this article, we have answered some frequently asked questions about average velocity and velocity. We have discussed the difference between average velocity and velocity, how to calculate the average velocity of an object, and how to determine on what interval(s) of time the particle is moving to the right. We have also provided an example of how to calculate the average velocity of an object.