Velocity Functions with Calculus: Videos & Practice Problems

Velocity Functions with Calculus connects position function, velocity function, and motion through derivatives and integrals. If position is given as \(x(t)\), the instantaneous velocity is found from \(v(t)=\frac{dx}{dt}\) . This lets you evaluate velocity at a specific time, identify an initial velocity by using \(t=0\), and determine when an object comes to a stop by solving \(v(t)=0\).

When velocity is given and position is needed, work in reverse with an indefinite integral: \(x(t)=\int v(t)\\,dt + C\) . The integration constant is found from an initial condition such as a known position at a known time. To find displacement over a time interval, use the definite integral: \(\Delta x=\int_{t_i}^{t_f} v(t)\\,dt\) , which gives a value without adding \(C\).

Average & Instantaneous Velocity from Position Function

When analyzing motion, the position function, denoted as x(t), represents an object's location at any given time t. By substituting a specific time value into this function, you can determine the exact position of the object at that moment. For example, if the position function is x(t) = t² + 4, then at t = 3 seconds, the position is calculated as 3² + 4 = 13 meters.

Velocity, which describes how fast an object’s position changes over time, can be understood in two ways: average velocity and instantaneous velocity. Average velocity is calculated over a time interval and is defined as the change in position divided by the change in time. Mathematically, average velocity v_avg between times t_i and t_f is given by the formula:

\[v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x(t_f) - x(t_i)}{t_f - t_i}\]

For instance, using the position function x(t) = t² + 4, to find the average velocity between t = 3 seconds and t = 5 seconds, first calculate the positions at these times: x(3) = 13 meters and x(5) = 29 meters. Then, the average velocity is:

\[v_{\text{avg}} = \frac{29 - 13}{5 - 3} = \frac{16}{2} = 8 \text{ meters per second}\]

Instantaneous velocity, on the other hand, refers to the velocity at a specific moment in time. It is found by taking the derivative of the position function with respect to time, which gives the velocity function v(t). The derivative represents the rate of change of position and is expressed as:

\[v(t) = \frac{d}{dt} x(t)\]

Using the example position function x(t) = t² + 4, the derivative is calculated by applying the power rule of differentiation, which states that for a function t^n, the derivative is n t^{n-1}. The derivative of a constant is zero. Therefore:

\[v(t) = \frac{d}{dt} (t^2 + 4) = 2t + 0 = 2t\]

To find the instantaneous velocity at t = 5 seconds, substitute 5 into the velocity function:

\[v(5) = 2 \times 5 = 10 \text{ meters per second}\]

In summary, average velocity measures the overall change in position over a time interval, while instantaneous velocity captures the exact rate of change of position at a single point in time. Calculating average velocity involves using the difference quotient, whereas instantaneous velocity requires differentiation of the position function.

The position function for a particle moving on the $x$-axis is $x\left(t\right)=-2t^2+18t+7$. What is the particle's initial velocity?

• A. $v_0=18\mathrm{m/s}$
• B. $v_0=7\mathrm{m/s}$
• C. $v_0=9\mathrm{m/s}$
• D. $v_0=4.5\mathrm{m/s}$

Answer: (A)

Average & Instantaneous Velocity from Position Function Example 1

To determine the position of a particle at the moment it comes to a stop, we start by understanding that the particle's velocity must be zero at that instant. The position function, denoted as \( x(t) \), gives the particle's location at any time \( t \), but to find the specific time when the particle stops, we need to analyze the velocity function \( v(t) \). Velocity is the derivative of the position function with respect to time, so by differentiating \( x(t) \), we obtain \( v(t) = \frac{dx}{dt} \).

For example, if the position function is given by \( x(t) = -3t^2 + 12t - 2.6 \), then the velocity function is found by differentiating each term:

\[v(t) = \frac{d}{dt}(-3t^2 + 12t - 2.6) = -6t + 12\]

Setting the velocity equal to zero to find when the particle stops, we solve the equation:

\[-6t + 12 = 0\]

Rearranging gives:

\[-6t = -12 \implies t = 2 \text{ seconds}\]

This means the particle stops momentarily at \( t = 2 \) seconds. To find the position at this time, substitute \( t = 2 \) back into the original position function:

\[x(2) = -3(2)^2 + 12(2) - 2.6 = -3(4) + 24 - 2.6 = -12 + 24 - 2.6 = 9.4 \text{ meters}\]

Thus, the particle's position when it comes to a stop is 9.4 meters. This process highlights the importance of understanding the relationship between position, velocity, and time, and how derivatives are used to analyze motion in calculus-based physics problems.

The position of a small object is given by $x=0.6e^{2t}-1$, where $t$ is in seconds and $x$ in meters. At what time is the small object travelling with a velocity $v=65.5\text{ m/s}$?

• A. $2.35\text{ s}$
• B. $2.0\text{ s}$
• C. $2.69\text{ s}$
• D. Never

Answer: (B)

Position & Displacement from Velocity Function

Understanding the relationship between position, velocity, and displacement is fundamental in kinematics. When given a position function x(t), the velocity function v(t) can be found by taking the derivative of the position with respect to time, expressed mathematically as \(v(t) = \frac{dx}{dt}\). For example, if the position function is \(x(t) = 2t^2 + 5\), differentiating term-by-term yields the velocity function \(v(t) = 4t\), since the derivative of \$2t^2\( is \)4t\( and the derivative of a constant is zero.

Conversely, when the velocity function is known and the goal is to determine the position function, integration is used. This process involves calculating the indefinite integral of the velocity function, which is the reverse operation of differentiation. The indefinite integral of \)v(t)\( is given by:

\[x(t) = \int v(t) \, dt + C\]

Here, \)C\( represents the integration constant, an unknown value that must be determined using initial conditions. For instance, if \)v(t) = 4t\(, integrating yields:

\[x(t) = \int 4t \, dt = 2t^2 + C\]

To find \)C\(, an initial condition such as the position at a specific time is required. If at \)t=0\(, the position is \)x=5\(, substituting these values into the equation gives:

\[5 = 2(0)^2 + C \implies C = 5\]

Thus, the complete position function is \)x(t) = 2t^2 + 5\(. This demonstrates how integration and differentiation are inverse operations, allowing one to move between velocity and position functions.

When calculating displacement, which is the change in position over a time interval, the definite integral of the velocity function is used. The displacement \(\Delta x\) from time \)t_i\( to \)t_f\( is given by:

\[\Delta x = \int_{t_i}^{t_f} v(t) \, dt = x(t_f) - x(t_i)\]

Unlike the indefinite integral, the definite integral does not include the integration constant because it evaluates the net change over the interval. For example, to find the displacement from \)t=1\( to \)t=4\( seconds for \)v(t) = 4t$, compute:

\[\Delta x = \int_1^4 4t \, dt = \left[ 2t^2 \right]_1^4 = 2(4)^2 - 2(1)^2 = 32 - 2 = 30\]

This result indicates a displacement of 30 meters over the specified time interval.

In summary, the indefinite integral of velocity yields the position function and requires an integration constant determined by initial conditions, while the definite integral of velocity over a time interval provides the displacement without the need for an integration constant. Mastery of these concepts enables solving a wide range of problems involving motion and kinematics.

A ball's velocity is given by $v\left(t\right)=\frac{1}{2}{t}^3-2t$. Calculate the ball's displacement from $t=1.5s$ to $t=3s$.

• A. $8.81m$
• B. $2.74m$
• C. $35.0m$
• D. $-2.74m$

Answer: (B)

Position & Displacement from Velocity Function Example 2

When given a velocity function and asked to find the position of an object at a specific time, the key step is to integrate the velocity function to obtain the position function. The position function, denoted as \( x(t) \), is found by calculating the indefinite integral of the velocity function and including an integration constant \( C \) to account for initial conditions.

For example, if the velocity function is \( v(t) = \cos t - 3t^2 \), the position function is determined by integrating each term separately:

\[x(t) = \int (\cos t - 3t^2) \, dt = \sin t - t^3 + C\]

Here, the integral of \( \cos t \) is \( \sin t \), and the integral of \( -3t^2 \) is \( -t^3 \) after simplifying the constants. The constant \( C \) represents the initial position offset and must be found using given initial conditions.

To find \( C \), use the known position at a specific time. For instance, if at \( t = 2 \), the position \( x = 10 \), substitute these values into the position function:

\[10 = \sin 2 - 2^3 + C\]

Calculating \( \sin 2 \) (in radians) and \( 2^3 = 8 \), we get:

\[10 = \sin 2 - 8 + C\]

Evaluating \( \sin 2 \approx 0.9093 \), the equation becomes:

\[10 = 0.9093 - 8 + C \implies C = 10 - 0.9093 + 8 = 17.0907\]

Thus, the position function is:

\[x(t) = \sin t - t^3 + 17.1\]

Finally, to find the position at \( t = 1.4 \), substitute \( t = 1.4 \) into the position function:

\[x(1.4) = \sin 1.4 - (1.4)^3 + 17.1\]

Calculating \( \sin 1.4 \approx 0.9854 \) and \( (1.4)^3 = 2.744 \), the position is:

\[x(1.4) = 0.9854 - 2.744 + 17.1 = 15.34\]

This result indicates the object’s position at \( t = 1.4 \) seconds is approximately 15.3 meters. Remember, when working with trigonometric functions in calculus problems, always ensure your calculator is set to radians mode to avoid incorrect results.

Here's what students ask on this topic:

How do you calculate instantaneous velocity from a position function using calculus?

To calculate instantaneous velocity from a position function $x\left(t\right)$, you need to find the derivative of the position function with respect to time. This derivative is the velocity function $v\left(t\right)=\frac{d}{d}x\left(t\right)$. The derivative gives the rate of change of position at any instant. For example, if $x\left(t\right)=t^2+\; 4$, then $v\left(t\right)=\frac{d}{d}(t^2+\; 4)\; =\; 2t$. To find the instantaneous velocity at a specific time, say $t=\; 5$, substitute into the velocity function: $v(5)\; =\; 2\; imes\; 5\; =\; 10$ meters per second. This method provides the velocity at a precise moment rather than over an interval.

What is the difference between average velocity and instantaneous velocity in calculus?

Average velocity over a time interval is calculated as the change in position divided by the change in time, expressed as $v{\mathrm{avg}}_{}=\frac{\mathrm{\backslash (\backslash Delta\backslash )\; x}}{\mathrm{\backslash (\backslash Delta\backslash )\; t}}=\frac{x}{f_{}}\frac{t}{f_{}}$. It represents the average rate of change of position between two times. Instantaneous velocity, on the other hand, is the velocity at a specific instant and is found by taking the derivative of the position function: $v(t)\; =\frac{d}{d}x(t)$. While average velocity gives an overall rate between two points, instantaneous velocity gives the exact rate at a single moment. For example, if $x(t)\; =\; t^2\; +\; 4$, average velocity between $t=3$ and $t=5$ is $(29\; -\; 13)\; /\; (5\; -\; 3)\; =\; 8$ m/s, but instantaneous velocity at $t=5$ is $2\; imes\; 5\; =\; 10$ m/s.

How do you find the position function from a given velocity function using calculus?

To find the position function $x(t)$ from a given velocity function $v(t)$, you perform the indefinite integral of the velocity function with respect to time: $x(t)\; =\; \backslash (\backslash int\backslash )\; v(t)\; \backslash ,\; dt\; +\; C$. The constant $C$ is the integration constant and represents the initial position. To determine $C$, you use an initial condition such as the position at a specific time, for example, $x(0)\; =\; x\_0$. For instance, if $v(t)\; =\; 4t$, then integrating gives $x(t)\; =\; 2t^2\; +\; C$. If the initial position at $t=0$ is 5 meters, then $x(0)\; =\; 2\; imes\; 0^2\; +\; C\; =\; 5$, so $C\; =\; 5$. Thus, the position function is $x(t)\; =\; 2t^2\; +\; 5$. This process reverses differentiation to recover position from velocity.

What is the difference between definite and indefinite integrals when working with velocity and position?

When working with velocity and position, an indefinite integral is used to find the position function from a velocity function. It is written as $x(t)\; =\; \backslash (\backslash int\backslash )\; v(t)\; \backslash ,\; dt\; +\; C$, where $C$ is the integration constant determined by initial conditions. This integral produces a function representing position at any time. A definite integral, however, calculates displacement over a time interval and is written as $\backslash (\backslash Delta\backslash )\; x\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; v(t)\; \backslash ,\; dt$. It evaluates the integral between two time limits and gives a numerical value for the change in position without an integration constant. The definite integral answers "how far" the object has moved between two times, while the indefinite integral provides the full position function including initial position.

How do you calculate displacement from a velocity function over a time interval?

Displacement over a time interval from $t\_i$ to $t\_f$ is calculated using the definite integral of the velocity function: $\backslash (\backslash Delta\backslash )\; x\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; v(t)\; \backslash ,\; dt$. This integral sums the velocity over the interval to find the net change in position. For example, if $v(t)\; =\; 4t$, then $\backslash (\backslash Delta\backslash )\; x\; =\; \backslash (\backslash int\backslash )\_1^4\; 4t\; \backslash ,\; dt$. Evaluating the integral, $\backslash (\backslash int\backslash )\; 4t\; \backslash ,\; dt\; =\; 2t^2$, so $\backslash (\backslash Delta\backslash )\; x\; =\; 2(4)^2\; -\; 2(1)^2\; =\; 32\; -\; 2\; =\; 30$ meters. This value represents the displacement, or how far the object has moved from $t=1$ to $t=4$. Note that no integration constant is needed for definite integrals.