2D Acceleration Functions with Calculus: Videos & Practice Problems

In 2D Acceleration Functions with Calculus, motion in the plane is connected through position, velocity, and acceleration using vector derivatives and integrals. Moving from position to velocity to acceleration means differentiating: \( \vec a(t)=\frac{d\vec v}{dt} \) . In unit-vector form, the \(i\) and \(j\) terms are the x- and y-components, so each component is differentiated separately.

Going backward from acceleration to velocity, or from velocity to position, means integrating. To find a full function, use an indefinite integral and include an integration constant; to find a change in velocity or displacement, use a definite integral over time. The key relationships are \( \Delta \vec v=\int_{t_i}^{t_f}\vec a(t)\\,dt \) and \( \Delta \vec r=\int_{t_i}^{t_f}\vec v(t)\\,dt \) . When needed, the magnitude of a 2D acceleration vector comes from its x- and y-components.

Solving Problems with Velocity and Acceleration Functions in 2D

Understanding the relationships between position, velocity, and acceleration functions is essential in analyzing motion in two dimensions. These three variables—position r(t), velocity v(t), and acceleration a(t)—are interconnected through derivatives and integrals, allowing you to move between them depending on the given information and what you need to find.

When moving from position to velocity or velocity to acceleration, you apply derivatives. For example, velocity is the first derivative of position with respect to time, expressed as \(v(t) = \frac{dr}{dt}\), and acceleration is the derivative of velocity, \(a(t) = \frac{dv}{dt}\). These derivatives can be average rates of change over intervals or instantaneous rates, which are the exact derivatives at a point.

Conversely, when moving backward—from acceleration to velocity or velocity to position—you use integrals. Integrating acceleration over time gives velocity, and integrating velocity gives position. There are two types of integrals to consider: indefinite integrals, which include an integration constant (denoted as +C), and definite integrals, which calculate the change in the function over a specific time interval without the constant. For example, the change in velocity \(\Delta v\) over time from \(t_i\) to \(t_f\( is found by the definite integral:

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

In practical problems, if you start with a position function such as \)r(t) = t^3 - 3t^2 + 4t \, \mathbf{i} + (3t^2 - 7t) \, \mathbf{j}\), you can find the acceleration by differentiating twice. First, differentiate \(r(t)\( to get velocity:

\[v(t) = \frac{dr}{dt} = (3t^2 - 6t + 4) \, \mathbf{i} + (6t - 7) \, \mathbf{j}\]

Then differentiate \)v(t)\( to find acceleration:

\[a(t) = \frac{dv}{dt} = (6t - 6) \, \mathbf{i} + 6 \, \mathbf{j}\]

Alternatively, if given an acceleration function like \)a(t) = (t - 3) \, \mathbf{i} + 7 \, \mathbf{j}\) and asked to find the change in velocity between two times, say \(t=6\) and \(t=8\), you compute the definite integral of acceleration over that interval:

\[\Delta v = \int_6^8 \left( (t - 3) \, \mathbf{i} + 7 \, \mathbf{j} \right) dt = \left[ \frac{t^2}{2} - 3t \right]_6^8 \mathbf{i} + \left[ 7t \right]_6^8 \mathbf{j}\]

Evaluating this gives:

\[\Delta v = \left( \frac{64}{2} - 24 - \left( \frac{36}{2} - 18 \right) \right) \mathbf{i} + (56 - 42) \mathbf{j} = 8 \, \mathbf{i} + 14 \, \mathbf{j}\]

Remember, when working with vectors, subtract components separately for each direction. The magnitude of this change in velocity vector can be found using the Pythagorean theorem:

\[|\Delta v| = \sqrt{(8)^2 + (14)^2} = \sqrt{64 + 196} = \sqrt{260} \approx 16.12\]

Mastering these derivative and integral relationships between position, velocity, and acceleration functions enables you to analyze motion comprehensively. Whether you are given one function and need to find another or calculate changes over time, applying these calculus concepts with attention to vector components is key to solving two-dimensional motion problems effectively.

A toy airplane is flying in an $\text{x/y}$ plane parallel to the ground with a velocity $v\left(t\right)=(4t-0.015t^3)\hat{ı}+(1.8+0.5t)\hat{ȷ}$. What is the magnitude of the airplane's acceleration at $t=30s$?

• A. $36.5\text{ m/}{s}^2$
• B. $285\text{ m/}{s}^2$
• C. $-36.0\text{ m/}{s}^2$
• D. $-268.2\text{ m/}{s}^2$

Answer: (A)

An iceboat on a frozen lake has acceleration . At $t=4s$, the iceboat's velocity is . What is the iceboat's displacement between $t=2s$ and $t=10s$?

• A.
• B.
• C.

Answer: (C)

Here's what students ask on this topic:

How do you find the acceleration function from a given 2D position function using calculus?

To find the acceleration function from a given 2D position function $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)}$, you need to differentiate twice with respect to time. First, take the derivative of the position function to get the velocity function $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)\; =\; \backslash (\backslash frac\{d\backslash vec\{r\}\backslash )\}\{dt\}}$. Then, differentiate the velocity function to obtain the acceleration function $\mathrm{\backslash (\backslash vec\{a\}\backslash )(t)\; =\; \backslash (\backslash frac\{d\backslash vec\{v\}\backslash )\}\{dt\}\; =\; \backslash (\backslash frac\{d^2\backslash vec\{r\}\backslash )\}\{dt^2\}}$. Each component (i.e., the $i$ and $j$ unit vectors) is differentiated separately. For example, if $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)\; =\; x(t)\backslash (\backslash hat\{i\}\backslash )\; +\; y(t)\backslash (\backslash hat\{j\}\backslash )}$, then $\mathrm{\backslash (\backslash vec\{a\}\backslash )(t)\; =\; \backslash (\backslash frac\{d^2x\}\{dt^2\}\backslash [\backslash hat\{i\}\backslash )\; +\; \backslash (\backslash frac\{d^2y\}\{dt^2\}\backslash ]\backslash hat\{j\}\backslash )}$. This process uses standard calculus derivatives and is essential for analyzing motion in two dimensions.

What is the difference between definite and indefinite integrals when calculating velocity or position from acceleration in 2D motion?

When calculating velocity or position from acceleration in 2D motion, the choice between definite and indefinite integrals depends on what you want to find. An indefinite integral is used to find the general function (velocity or position) and includes an integration constant $C$. For example, integrating acceleration $\mathrm{\backslash (\backslash vec\{a\}\backslash )(t)}$ gives velocity $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)\; =\; \backslash (\backslash int\backslash )\; \backslash (\backslash vec\{a\}\backslash )(t)\; dt\; +\; \backslash (\backslash vec\{C\}\backslash )}$. The constant is determined by initial conditions. A definite integral calculates the change in velocity or displacement over a time interval $[t\_i,\; t\_f]$ without an integration constant: $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{v\}\backslash )\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; \backslash (\backslash vec\{a\}\backslash )(t)\; dt}$ or $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{r\}\backslash )\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; \backslash (\backslash vec\{v\}\backslash )(t)\; dt}$. Definite integrals give the net change between two times, which is useful for problems asking for displacement or change in velocity.

How do you calculate the change in velocity vector from a given 2D acceleration function over a time interval?

To calculate the change in velocity vector $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{v\}\backslash )}$ from a given 2D acceleration function $\mathrm{\backslash (\backslash vec\{a\}\backslash )(t)}$ over a time interval $[t\_i,\; t\_f]$, you perform a definite integral of the acceleration function with respect to time: $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{v\}\backslash )\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; \backslash (\backslash vec\{a\}\backslash )(t)\; dt}$. This integral is evaluated component-wise. For example, if $\mathrm{\backslash (\backslash vec\{a\}\backslash )(t)\; =\; a\_x(t)\backslash (\backslash hat\{i\}\backslash )\; +\; a\_y(t)\backslash (\backslash hat\{j\}\backslash )}$, then $\mathrm{\backslash (\backslash Delta\backslash )\; v\_x\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; a\_x(t)\; dt}$ and $\mathrm{\backslash (\backslash Delta\backslash )\; v\_y\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; a\_y(t)\; dt}$. After integrating, substitute the upper and lower limits and subtract to find the net change in each component. The resulting vector $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{v\}\backslash )\; =\; \backslash (\backslash Delta\backslash )\; v\_x\; \backslash (\backslash hat\{i\}\backslash )\; +\; \backslash (\backslash Delta\backslash )\; v\_y\; \backslash (\backslash hat\{j\}\backslash )}$ represents the total change in velocity over the time interval.

What is the process to go from velocity to position in 2D motion using calculus?

To go from velocity $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)}$ to position $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)}$ in 2D motion, you integrate the velocity function with respect to time. This is done by calculating the indefinite integral: $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)\; =\; \backslash (\backslash int\backslash )\; \backslash (\backslash vec\{v\}\backslash )(t)\; dt\; +\; \backslash (\backslash vec\{C\}\backslash )}$, where $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$ is the integration constant vector determined by initial position conditions. The integration is performed separately for each component: $\mathrm{r\_x(t)\; =\; \backslash (\backslash int\backslash )\; v\_x(t)\; dt\; +\; C\_x}$ and $\mathrm{r\_y(t)\; =\; \backslash (\backslash int\backslash )\; v\_y(t)\; dt\; +\; C\_y}$. If you want the displacement over a time interval, use a definite integral instead: $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{r\}\backslash )\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; \backslash (\backslash vec\{v\}\backslash )(t)\; dt}$. This process allows you to reconstruct the position function from velocity data in two dimensions.

How do you interpret the components of 2D acceleration vectors in terms of unit vectors?

In 2D motion, acceleration vectors are expressed in terms of unit vectors $\mathrm{\backslash (\backslash hat\{i\}\backslash )}$ and $\mathrm{\backslash (\backslash hat\{j\}\backslash )}$, which represent the x- and y-directions respectively. The acceleration vector is written as $\mathrm{\backslash (\backslash vec\{a\}\backslash )(t)\; =\; a\_x(t)\backslash (\backslash hat\{i\}\backslash )\; +\; a\_y(t)\backslash (\backslash hat\{j\}\backslash )}$, where $\mathrm{a\_x(t)}$ and $\mathrm{a\_y(t)}$ are the acceleration components along the x- and y-axes. Each component is treated independently when differentiating or integrating. For example, when differentiating velocity to get acceleration, you differentiate $\mathrm{v\_x(t)}$ to get $\mathrm{a\_x(t)}$ and $\mathrm{v\_y(t)}$ to get $\mathrm{a\_y(t)}$. This vector form allows you to analyze motion in two dimensions by considering how acceleration affects each direction separately.