Motion Analysis: Videos & Practice Problems

Derivatives are essential in understanding motion, where the first derivative of position gives velocity, and the second derivative provides acceleration. Velocity is the change in position over time, while acceleration is the change in velocity over time. Average acceleration is calculated as the change in velocity divided by the change in time, yielding units of meters per second squared. Instantaneous acceleration is found by differentiating the velocity function. This relationship between position, velocity, and acceleration is crucial for analyzing real-world motion scenarios.

Derivatives Applied To Velocity

Derivatives play a crucial role in understanding motion and velocity in real-world applications. When analyzing the motion of an object, we often describe its position and velocity as functions of time. The position function, denoted as \( s(t) \), indicates the location of the object at a specific time \( t \), while the velocity function represents how fast the object is moving at that time.

To illustrate this, consider a scenario where the position of an object is given by the function \( s(t) = t^2 \) (in meters). This function suggests that as time progresses, the object's position increases in a parabolic manner. For example, at \( t = 1 \) second, the object is at 1 meter; at \( t = 2 \) seconds, it reaches 4 meters; and at \( t = 3 \) seconds, it is at 9 meters. The relationship between position and velocity is defined by the derivative of the position function. Thus, the velocity \( v(t) \) can be calculated as the first derivative of the position function:

$$ v(t) = \frac{ds}{dt} = \frac{d(t^2)}{dt} = 2t $$

This indicates that the velocity function is linear, with a slope of 2, meaning the object's speed increases as time progresses.

To further analyze motion, we can calculate displacement, which is the change in position over a specified time interval. Displacement can be determined by subtracting the initial position from the final position. For a time interval from \( t = 0 \) to \( t = 2 \) seconds:

Final position at \( t = 2 \): \( s(2) = 2^2 = 4 \) meters

Initial position at \( t = 0 \): \( s(0) = 0^2 = 0 \) meters

Displacement = Final position - Initial position = \( 4 - 0 = 4 \) meters.

Next, we can differentiate between average velocity and instantaneous velocity. Average velocity is calculated over a specific time interval and is represented by the slope of the secant line connecting two points on the position graph. For our example, the average velocity over the interval from \( t = 0 \) to \( t = 2 \) seconds is:

Average velocity = Displacement / Change in time = \( \frac{4 \text{ meters}}{2 \text{ seconds}} = 2 \text{ m/s} \).

In contrast, instantaneous velocity is the velocity at a specific moment, represented by the slope of the tangent line at that point. To find the instantaneous velocity at \( t = 2 \) seconds, we substitute \( t = 2 \) into the velocity function:

Instantaneous velocity = \( v(2) = 2 \times 2 = 4 \text{ m/s} \).

Lastly, it is important to distinguish between speed and velocity. Speed is the absolute value of velocity and is always a non-negative quantity. In this case, since the instantaneous velocity is \( 4 \text{ m/s} \), the speed is also \( 4 \text{ m/s} \). However, if the velocity were negative, the speed would still be positive, reflecting the object's rate of motion without regard to direction.

In summary, by utilizing the position function and understanding the relationships between position, velocity, displacement, average velocity, instantaneous velocity, and speed, we can effectively analyze motion in various real-world contexts.

Given the position equation $s\left(t\right)$, calculate the average velocity (in meters per second) based on the given interval, and the instantaneous velocity (in meters per second) at the end of the time interval.
$s\left(t\right)=9t-t^2$, $0\le t\le3$

• A. Average: $v_{avg}=6\frac{m}{s}$, Instantaneous: $v\left(3\right)=3\frac{m}{s}$
• B. Average: $v_{avg}=10\frac{m}{s}$ , Instantaneous: $v\left(3\right)=18\frac{m}{s}$
• C. Average: $v_{avg}=18\frac{m}{s}$ , Instantaneous: $v\left(3\right)=10\frac{m}{s}$
• D. Average: $v_{avg}=6\frac{m}{s}$ , Instantaneous: $v\left(3\right)=0$

Answer: (A)

Given the position equation $s\left(t\right)$ , calculate the average velocity (in meters per second) based on the given time interval, and the instantaneous velocity (in meters per second) at the end of the time interval.
$s\left(t\right)=\frac{30}{t+5}$, $-4\le t\le0$

• A. Average: $v_{avg}=-6\frac{m}{s}$, Instantaneous: $v\left(0\right)=0$
• B. Average: $v_{avg}=-6\frac{m}{s}$, Instantaneous: $v\left(0\right)=30\frac{m}{s}$
• C. Average: $v_{avg}=6\frac{m}{s}$ , Instantaneous: $v\left(0\right)=18\frac{m}{s}$
• D. Average: $v_{avg}=-6\frac{m}{s}$, Instantaneous: $v\left(0\right)=-1.2\frac{m}{s}$

Answer: (D)

Derivatives Applied To Velocity Example 1

Understanding the relationship between position and velocity is crucial in solving real-world motion problems, particularly those involving projectiles. When analyzing a projectile shot vertically upward, we can express its height as a function of time. For example, the height \( h(t) \) of a projectile launched from a height of 5 meters with an initial velocity of 20 meters per second can be modeled by the equation:

\[ h(t) = 5 + 20t - 4.9t^2 \]

To find the vertical velocity at a specific time, we utilize the concept that velocity is the first derivative of the position function. Thus, we differentiate the height function:

\[ h'(t) = \frac{d}{dt}(5 + 20t - 4.9t^2) = 0 + 20 - 9.8t = 20 - 9.8t \]

To determine the vertical velocity at \( t = 3 \) seconds, we substitute \( t \) into the velocity function:

\[ h'(3) = 20 - 9.8(3) = 20 - 29.4 = -9.4 \, \text{m/s} \]

This negative value indicates that the projectile is moving downward at that moment. The units of velocity are meters per second, consistent with the height being measured in meters and time in seconds.

Next, to find when the projectile reaches its maximum height, we recognize that this occurs when the vertical velocity is zero. Setting the velocity function equal to zero gives:

\[ 20 - 9.8t = 0 \]

Solving for \( t \) yields:

\[ 9.8t = 20 \quad \Rightarrow \quad t = \frac{20}{9.8} \approx 2.04 \, \text{seconds} \]

Now that we have the time at which the maximum height occurs, we can find the maximum height by substituting this time back into the original height equation:

\[ h(2.04) = 5 + 20(2.04) - 4.9(2.04^2) \]

Calculating this gives:

\[ h(2.04) \approx 5 + 40.8 - 4.9(4.1616) \approx 5 + 40.8 - 20.4 \approx 25.41 \, \text{meters} \]

Thus, the maximum height reached by the projectile is approximately 25.41 meters. This process illustrates how to apply calculus concepts, such as derivatives, to solve practical problems involving motion and projectile trajectories.

Derivatives Applied To Acceleration

In the study of motion, understanding acceleration is crucial as it describes how velocity changes over time. Acceleration can be mathematically defined as the first derivative of the velocity function with respect to time, or equivalently, as the second derivative of the position function. This relationship highlights that if you start with a position function, taking the first derivative yields velocity, and taking the derivative of that result gives you acceleration.

To calculate acceleration, you can use the formula:

$$ a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} $$

where \( a(t) \) is acceleration, \( v \) is velocity, and \( s \) is position.

When solving problems involving acceleration, it is essential to identify the type of function provided. For instance, if given a velocity function, you can find the change in velocity over a specified time interval by subtracting the initial velocity from the final velocity. This is expressed as:

$$ \Delta v = v(t_f) - v(t_i) $$

where \( t_f \) is the final time and \( t_i \) is the initial time.

Average acceleration can be calculated similarly to average velocity, defined as the change in velocity divided by the change in time:

$$ \text{Average Acceleration} = \frac{\Delta v}{\Delta t} $$

where \( \Delta t = t_f - t_i \). The units for average acceleration are typically expressed in meters per second squared (m/s²), indicating a distance per squared time.

For instantaneous acceleration, which reflects the acceleration at a specific moment, you derive the acceleration function from the velocity function. Using the power rule for differentiation, you can find the acceleration function and then evaluate it at the desired time point. For example, if the velocity function is given as:

$$ v(t) = t^3 - 3t^2 + 2t $$

the acceleration function can be derived as:

$$ a(t) = \frac{d}{dt}(t^3 - 3t^2 + 2t) = 3t^2 - 6t + 2 $$

To find the instantaneous acceleration at a specific time, substitute that time into the acceleration function. For instance, to find the instantaneous acceleration at \( t = 3 \) seconds:

$$ a(3) = 3(3^2) - 6(3) + 2 = 27 - 18 + 2 = 11 \text{ m/s}^2 $$

In summary, understanding the relationships between position, velocity, and acceleration through derivatives allows for the analysis of motion in various contexts, providing a mathematical framework to describe how objects move and change speed over time.

Given the position of an object $s\left(t\right)=12t-t^2$ (in meters), find the acceleration of the object at $t=5$ seconds.

• A. Instantaneous Acceleration $a=9.8\frac{m}{s^2}$
• B. Instantaneous Acceleration $a=2\frac{m}{s^2}$
• C. Instantaneous Acceleration $a=-2\frac{m}{s^2}$
• D. Instantaneous Acceleration $a=0\frac{m}{s^2}$

Answer: (C)

Given below is the graph of velocity with respect to time. At which time(s) would acceleration be 0?

• A. At $t=0$
• B. At $t=1$ & $t=3$
• C. At $t=0$ & $t=5$
• D. At $t=1$, $t=2$ & $t=3$

Answer: (D)

Derivatives Applied To Acceleration Example 2

In projectile motion, the height of an object shot vertically upward can be modeled by a quadratic equation. For example, if the height \( h(t) \) in meters is given by the equation:

\[ h(t) = 15t - 4.9t^2 \]

where \( t \) is the time in seconds, we can derive important information about the motion of the projectile, such as its vertical velocity and acceleration.

To find the vertical velocity at a specific time, we need to calculate the first derivative of the height function with respect to time. The derivative of the height function is:

\[ v(t) = \frac{dh}{dt} = 15 - 9.8t \]

To find the vertical velocity at \( t = 3 \) seconds, we substitute \( t \) into the velocity equation:

\[ v(3) = 15 - 9.8 \times 3 = 15 - 29.4 = -14.4 \, \text{m/s} \]

This negative value indicates that the projectile is moving downward at that time.

Next, to determine the vertical acceleration, we take the second derivative of the height function, which is also the first derivative of the velocity function:

\[ a(t) = \frac{dv}{dt} = -9.8 \, \text{m/s}^2 \]

This constant value represents the acceleration due to gravity, which remains the same regardless of the time since gravity continuously acts on the projectile.

To find when the projectile reaches half of its maximum height, we first calculate half of the maximum height. Given that the maximum height is 11.5 meters, half of this height is:

\[ \frac{11.5}{2} = 5.75 \, \text{meters} \]

We set the height equation equal to 5.75 meters to find the time:

\[ 15t - 4.9t^2 = 5.75 \]

Rearranging gives us a quadratic equation:

\[ -4.9t^2 + 15t - 5.75 = 0 \]

Multiplying through by -1 simplifies it to:

\[ 4.9t^2 - 15t + 5.75 = 0 \]

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -15 \), and \( c = 5.75 \), we can find the values of \( t \). Plugging in these values yields:

\[ t = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 4.9 \cdot 5.75}}{2 \cdot 4.9} \]

Calculating this gives two possible times: approximately \( t \approx 0.45 \) seconds and \( t \approx 2.61 \) seconds. The first time, \( 0.45 \) seconds, represents when the projectile first reaches half of its maximum height, while the second time corresponds to when it returns to that height on its way down.

In summary, understanding the relationships between position, velocity, and acceleration through derivatives allows us to analyze projectile motion effectively. The key concepts include recognizing the role of gravity, applying the quadratic formula, and interpreting the results in the context of the motion of the projectile.

Here’s what students ask on this topic:

What is the relationship between position, velocity, and acceleration in motion analysis?

The relationship between position, velocity, and acceleration is fundamental in motion analysis. The position function, s(t), describes the location of an object at a given time t. The first derivative of the position function with respect to time, ds/dt, gives the velocity function, v(t), which indicates how fast the object is moving. The second derivative of the position function, d²s/dt², or the first derivative of the velocity function, dv/dt, provides the acceleration function, a(t), which describes how the velocity of the object changes over time. This relationship is crucial for understanding and analyzing real-world motion scenarios.

How do you calculate average velocity and instantaneous velocity?

Average velocity is calculated by dividing the change in position by the change in time over a specific interval. Mathematically, it is expressed as:

ΔsΔt

where Δs is the change in position and Δt is the change in time. Instantaneous velocity, on the other hand, is the velocity at a specific point in time and is found by taking the derivative of the position function with respect to time. If s(t) is the position function, then the instantaneous velocity v(t) is:

dsdt

What is the difference between speed and velocity?

Speed and velocity are related but distinct concepts in motion analysis. Speed is a scalar quantity that represents the magnitude of how fast an object is moving, regardless of direction. It is always positive. Velocity, on the other hand, is a vector quantity that includes both the speed and the direction of the object's motion. Velocity can be positive or negative depending on the direction. For example, if an object moves 5 meters per second to the right, its speed is 5 m/s, and its velocity is +5 m/s. If it moves to the left at the same speed, its velocity is -5 m/s.

How do you find the acceleration function from a given velocity function?

To find the acceleration function from a given velocity function, you need to take the first derivative of the velocity function with respect to time. If v(t) is the velocity function, then the acceleration function a(t) is given by:

dvdt

This derivative represents the rate of change of velocity over time. For example, if the velocity function is v(t) = 3t² - 6t + 2, the acceleration function a(t) would be found by differentiating v(t) with respect to t, resulting in a(t) = 6t - 6.

What is the significance of the units of acceleration?

The units of acceleration are significant because they describe how quickly the velocity of an object is changing over time. Acceleration is typically measured in meters per second squared (m/s²). This unit indicates that the velocity of an object is changing by a certain number of meters per second every second. For example, an acceleration of 2 m/s² means that the object's velocity increases by 2 meters per second every second. Understanding these units is crucial for analyzing and interpreting motion in real-world scenarios.