# Calculus 1: Motion Problem Examples

TLDRThis educational video script discusses solving motion problems by integrating the concepts of derivatives and integrals. The instructor begins by explaining how to find the displacement function X(t) from a given velocity function V(t), using integration and applying initial conditions to solve for the constant of integration. The script continues with deriving the acceleration function from the velocity function, showcasing the process of differentiation. Examples include solving for displacement with a velocity function of -32t + 20 and acceleration with a cosine function, where the instructor also corrects a minor mistake regarding the sign of the derivative. The video concludes with a prompt for viewers to visit the instructor's website for more calculus problem-solving examples.

###### Takeaways

- ๐ข Motion problems involve both derivatives and integrals.
- ๐ To find displacement X(t), take the antiderivative of the velocity function V(t).
- ๐ Displacement formula: X(t) = โซ V(t) dt.
- ๐งฎ Example: Given V(t) = -32t + 20, X(t) is found by integrating V(t).
- ๐ Initial values help determine the constant of integration.
- โ๏ธ In the example, X(0) = 0 helps find C, resulting in X(t) = -16t^2 + 20t.
- ๐ Acceleration a(t) is the derivative of the velocity function.
- ๐ Example: For V(t) = 10, X(t) is found to be 10t + 10 using the initial value X(0) = 10.
- ๐ Another example: V(t) = cos(t) leads to X(t) = sin(t) - 5 given X(0) = -5.
- ๐งโ๐ซ Acceleration for V(t) = cos(t) is found to be -sin(t).

###### Q & A

### What is the relationship between derivatives and integrals in motion problems?

-In motion problems, derivatives and integrals are related as follows: derivatives are used to find the velocity and acceleration of an object, while integrals are used to find the displacement and the distance traveled by the object.

### What is the formula to find displacement in terms of velocity?

-The formula to find displacement (X of T) is the integral of the velocity function (V of T) with respect to time (DT), which means X of T is the antiderivative of V of T.

### How do you find the initial value of the displacement function given an initial condition?

-To find the initial value of the displacement function, you plug in the initial condition (e.g., X of 0) into the displacement function and solve for the constant of integration (C).

### What is the velocity function given in the script?

-The velocity function given in the script is V of T, which is -32 times one-half T squared plus 20 T.

### What is the displacement function X of T derived from the velocity function?

-The displacement function X of T is derived from the velocity function by taking the antiderivative, resulting in -16 T squared plus 20 T plus C.

### What is the acceleration function a of T?

-The acceleration function a of T is the derivative of the velocity function V of T. In the script, it is found to be -32.

### How do you find the constant of integration C in the displacement function?

-To find the constant of integration C, you use the initial condition where X of 0 equals zero. By plugging in 0 into the displacement function, you can solve for C.

### What is the process to find the displacement function X of T when given a velocity function V of T?

-The process involves taking the antiderivative of the velocity function V of T to get the displacement function X of T, and then using the initial condition to solve for the constant of integration.

### What is the correct antiderivative of the velocity function 10 T?

-The correct antiderivative of the velocity function 10 T is 5 T squared plus C, where C is the constant of integration.

### How do you find the acceleration function a of T when given a velocity function V of T?

-To find the acceleration function a of T, you take the derivative of the velocity function V of T. In the script, the derivative of cosine of T is found to be negative sine of T.

### What is the mistake made in the script regarding the derivative of the velocity function?

-The mistake in the script is the incorrect sign for the derivative of the velocity function. The correct derivative of cosine of T is negative sine of T, but the script incorrectly states it as positive sine of T.

### What is the correct displacement function if the initial condition is X of 0 equals negative five?

-The correct displacement function, given the initial condition X of 0 equals negative five, would be negative cosine T plus C, where C is found to be negative four.

### What is the final form of the acceleration function a of T in the script?

-The final form of the acceleration function a of T in the script is sine of T minus five.

### Where can one find more examples and step-by-step solutions for calculus problems?

-More examples and step-by-step solutions for calculus problems can be found on the speaker's website, which offers free access to over 400 calculus questions.

###### Outlines

##### ๐ Introduction to Motion Problems in Calculus

This paragraph introduces the topic of motion problems, which are common in calculus and involve both derivatives and integrals. The speaker explains the importance of understanding the relationship between these two concepts and how to solve problems in both directions. The main focus is on finding the displacement and acceleration of an object moving in a straight line, given its velocity function and an initial value.

###### Mindmap

###### Keywords

##### ๐กMotion Problems

##### ๐กDerivatives

##### ๐กIntegrals

##### ๐กVelocity Function

##### ๐กDisplacement

##### ๐กAntiderivatives

##### ๐กInitial Value

##### ๐กAcceleration

##### ๐กConstant of Integration

##### ๐กTrigonometric Functions

###### Highlights

Introduction to motion problems involving derivatives and integrals.

Understanding the relationship between derivatives and integrals in motion problems.

Finding displacement X of T using the integral of velocity function V of T.

Calculating X of T as the antiderivative of V of T.

Initial value problem setup with X of zero equals zero.

Determining the constant of integration C to be zero.

Deriving acceleration a of T from the velocity function.

Solving for X of T with a given initial condition X of 0 equals 10.

Finding the constant of integration C to be 10.

Using the antiderivative to find X of T when V of T is cosine of T.

Determining the constant of integration when X is zero.

Correcting the sign error in the derivative of velocity.

Final expression for acceleration a of T as sine of T minus five.

Derivative of cosine resulting in negative sine of T.

Resource recommendation for more calculus examples and solutions.

###### Transcripts

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