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In the real world, everything is in motion. Objects move either with constant or varying velocities. For example, in everyday experience, you noticed when you step on the gas pedal of your car or brake one, the car's speed increases or decreases, Or in a turn your car's direction changes.

These changes in the magnitude of velocity or direction of a moving body, in physics, are described by acceleration.

In other words, the average acceleration is defined as the rate of change of the object's velocity and is given by the following formula:

\[\text{Average Acceleration}\equiv\frac{\text{change in velocity}}{\text{time elapsed}}\]

Typically average acceleration is denoted by $\bar a$ or $\vec{a}_{\text{ave}}$.

The units for acceleration are meters per squared second ($\rm \dfrac{m}{s^2}$).

**Note $1$:** In above, since the change in velocity $\Delta v$ occurs at a finite time interval $\Delta t$ so we should define average acceleration. If time intervals become infinitesimal short, in that case, we have instantaneous acceleration or shortly acceleration and is given by the below formula

\[ \vec{a} = \lim_{\Delta t \to 0} \frac{\text{change in velocity}}{\text{elapsed time}\,\Delta t} \]

**Note $2$:** Velocity is a vector quantity which means it has magnitude and direction. If one of those changes over a period of time, then we have acceleration. An object with the uniform (constant) speed in a circular path can still have an acceleration like satellites around the Earth which move with constant but undergoes a type of acceleration called centripetal acceleration.

**Note $3$:** Change in velocity is defined as the difference between initial ($\vec{v_1}$) and final ($\vec{v_2}$) velocities as below \[\Delta\vec{v}=\vec{v_2}-\vec{v_1}\] **Note $4$:** If an object moves in a straight line since in this case there is no change in direction, the only part of the velocity vector changes is its magnitude. Thus, we can omit its direction sign i.e. $\vec v=v$.

**Note $5$: **The average acceleration vector points in the same direction as the vector $\Delta \vec{v}$.

By practicing these examples of average acceleration, you can learn how to find the average acceleration. For a thorough and complete explanation related to velocity and acceleration and their examples go to course pages.

**Example (1):** A car is traveling in a straight line along a highway at a constant speed of $80$ miles per hour for 10 seconds. Find its acceleration?

**Solution:** Average acceleration is a change in velocity divided by the time taken. Since the car's velocity (magnitude and direction) is constant over the entire course, so by definition of average acceleration, it is zero i.e. $\bar{a}=0$.

**Example (2):** A plane has a take-off speed of $300\,{\rm \frac{km}{h}}$. What is the average acceleration (in $\rm \frac{m}{s^2}$) of the plane if the plane started from rest and took 45 seconds to take off?

**Solution:** Plane is initially at rest so $\vec{v_1}=0$ and its take off speed is $\vec{v_2}=300\,{\rm \frac{km}{h}}$. First off, convert ${\rm \frac{km}{h}}$ to SI units of velocity ${\rm \frac{m}{s}}$ as below\begin{align*} \rm{300\,\frac{km}{h}}&=\rm{300\,\frac{1000\,m}{3600\,s}}\\\\&=\rm{300\,\frac{1000}{3600}\,\frac{m}{s}}\\\\ &= \rm{83.4\,\frac{m}{s}} \end{align*} Now ratio of change in velocity, $\Delta \vec{v}=83.4\,{\rm \frac ms}$ over time elapsed $\Delta t=45\,\rm s$ is definition of average acceleration.\begin{align*} \bar{a}&=\frac{\Delta \vec{v}}{\Delta t} \\\\ &=\frac{83.4\,\rm{m/s}}{40\,{\rm s}}=2.085\,\rm{\frac{m}{s^2}} \end{align*}

**Example (3):** What average acceleration is needed to accelerate a car from $36\,{\rm \frac{km}{h}}$ to $72\,{\rm \frac{km}{h}}$ in $25$ seconds?

**Solution:** Initial and final velocities are $36\,{\rm \frac{km}{h}}$ and $72\,{\rm \frac{km}{h}}$, respectively. As before, convert them in SI units as below \begin{align*} \rm{\frac{km}{h}} &= \rm{\frac{1000\,m}{3600\,s}}\\ \\&= \rm{\frac{1000}{3600}\,\frac{m}{s}}\\\\ &= \rm{\frac{10}{36}\,\frac{m}{s}} \end{align*} In other words, multiply them by $\frac{10}{36}$. Then, $v_1=36\cdot\frac{10}{36}=10\,{\rm \frac ms}$ and $v_2=72\cdot\frac{10}{36}=20\,{\rm \frac ms}$. Now dividing change of velocity, $\Delta v=20-10=10\,{\rm \frac ms}$ over the time elapsed $\Delta t=25\,{\rm s}$ we get the desired average acceleration \begin{align*} \bar{a}&=\frac{\Delta \vec{v}}{\Delta t} \\ &=\frac{20\,\rm{m/s}}{25\,{\rm s}}=0.8\,\rm{\frac{m}{s^2}} \end{align*}

**Example (4):** Starting with a constant velocity of $50\,{\rm \frac{km}{h}}$, a car accelerates for 32 seconds at an acceleration of $0.5\,{\rm \frac{m}{s^2}}$. What is the velocity of the car at the end of the period of $32$ seconds of acceleration?

**Solution:**

Here initial velocity, acceleration and time interval over which this car accelerates are given and the final velocity is requested. Therefore,

using definition of average acceleration we get, \begin{align*} \bar{a} &=\frac{\vec{v_2}-\vec{v_1}}{\Delta t} \\ 0.5 &= \frac{v_{2}-50\times\frac{10}{36}}{32}\\ 32\times 0.5 &= v_{2}-50\times\frac{10}{36}\\ \Rightarrow v_2 &=16+13.9=29.9\,{\rm \frac ms} \end{align*} For the second equality refer to Note $4$.

**Example (5):** How long does it take to accelerate a car from a speed of $50\,{\rm \frac{km}{h}}$ to a speed of $100\,{\rm \frac{km}{h}}$ at a rate of $1\,{\rm \frac{m}{s^2}}$?

**Solution: **In this example, the unknown quantity is the time interval over which average acceleration occurs. Thus from definition of average acceleration along a straight line, we have \begin{align*} \bar{a} &=\frac{\vec{v_2}-\vec{v_1}}{\Delta t} \\ 1&=\frac{\left(100-50\right)\times \frac{10}{36}}{t}\\ \Rightarrow t&=\frac{\left(100-50\right)\times \frac{10}{36}}{1}\\ &=13.9\,{\rm s} \end{align*} In the second equality, kilometer/hour ($km/h$) is converted into the meter/second ($m/s$) by multiplying by $\frac{10}{36}$.

It is also possible to compute the acceleration of a moving object using a position versus time graph.

**Example (6):**

A car start moving with a velocity of $-4\,{\rm m/s}$ from position $+4\,{\rm m}$. After $2\,{\rm s}$, its position is $-1\,{\rm m}$ with final velocity $-1\,{\rm m/s}$. Find:

(a) What is the displacement of the car?

(b) What is the average velocity of the car?

(c) What is the average acceleration of the car?

**Solution: **

(a) Displacement is defined as the change in position of an object \begin{align*}\Delta x&=x_2-x_1\\&=-1-4=-5\,{\rm m}\end{align*}

(b) Average velocity is $\bar{v}=\frac{\Delta x}{\Delta t}$ so \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\&=\frac{-5}{2}\\&=-2.5\,{\rm m/s}\end{align*}

(c) By applying average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$, we get \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\&=\frac{-1-(-4)}{2}=+1.5\,{\rm m/s^2}\end{align*}In this example, the negatives of velocities indicates the car moves in the negative $x$ direction but the average acceleration is toward the positive $x$ direction. Consequently the car is slowing down or in other words, because $a.v<0$ so the motion is slowing down.

All these kinematics quantities are also found using a position vs. time graph.

**Example (7):**

A $5-g$ ball is dropped from a height. It bounces off a brick wall at $10\,{\rm m/s}$ and rebounds at $8\,{\rm m/s}$. The ball is in contact with the wall for $2\,{\rm ms}$. Find the average acceleration of the ball during the contact time?

**Solution:** Let the positive direction be up so the given data are $v_1=-10\,{\rm m/s}$, $v_2=+8\,{\rm m/s}$ and $\Delta t=2\times 10^{-3}\,{\rm s}$ which we converted the milliseconds into seconds.

By applying average acceleration formula, we get \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{+8-(-10)}{2\times 10^{-3}}\\\\&=9\times 10^{3}\,{\rm m/s^2}\end{align*}

**Example (8):**

A plastic bullet of $2\,{\rm g}$ from a rifle is fired to a brick wall. It strikes the wall with $20\,{\rm m/s}$ and returns at $18\,{\rm m/s}$. If the contact time of the ball with the wall is $3\,{\rm ms}$, what is the bullet's average acceleration?

**Solution:** suppose the positive direction be toward the positive $x$ direction. Thus, we have $v_1=+20\,{\rm m/s}$, $v_2=-18\,{\rm m/s}$ and $\Delta t=3\times 10^{-3}\,{\rm s}$. Therefore, the average acceleration definition gives \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{-18-(+20)}{3\times 10^{-3}}\\\\&=-12\times 10^{3}\,{\rm m/s^2}\end{align*} The minus sign of acceleration indicates its direction which is toward the negative $x$ axis.

**Example (9): The velocity-time graph of a moving car along a straight path is shown below. Find its average acceleration during the first 2 seconds of motion.**

**Solution**: the slope of a line segment connecting two points on a velocity-time graph gives us the average acceleration.

The slope is also defines as change in vertical axis divided by a change in horizontal axis. The starting point has coordinate $(v=-4\,{\rm m/s},t=0)$ and final point is $(v=0,t=2\,{\rm s})$. So, the slope of line joining these points is found as below \begin{align*}\text{slope=acceleration}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{0-(-4)}{2-0}\\\\&=+2\quad {\rm m/s^2}\end{align*} Thus, the car increases its speed at a constant rate of $2\,{\rm m/s^2}$ in 2 s.

Velocity is a vector quantity with a magnitude and direction. If one of those changes during a finite period of time $\Delta t$, then we have average acceleration (vector) and is given by the following formula: \begin{align*} \vec{a}_\text{ave}&=\frac{\text{change in velocity vector}}{\text{elapsed time}}\\ &=\frac{\Delta \vec{v}}{\Delta t}\\ &=\frac{\vec{v}_2-\vec{v}_1}{t_2-t_1} \end{align*} SI unit of acceleration is meter per second per second ($m/s^2$) and its dimension is ${\rm L\,T^{-2}}$.

Practicing more problems on speed, velocity, and acceleration are also helped a better understand of the average acceleration concept.

**Last Update : 9/9/2020**

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