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Average Acceleration Definition and solved examples for secondary school

Average Acceleration Definition and solved examples for secondary school

 

Definition of acceleration:

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, described by acceleration.

In other words, acceleration measures the rate of change of the object's velocity and is given by the following formula:
\[\text{Average Acceleration}\equiv\frac{\text{change in velocity}}{\text{elapsed time}}\]
Typically average acceleration is denoted by $\bar a$ or $\vec{a}_{\text{ave}}$.

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

Change in velocity of a moving car divided by the elapsed time is shown.

Some quick facts about acceleration:

  •  In above, since the change in velocity 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 and is given by below formula

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

  •  Remember that the 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 uniform (constant) speed in the 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. 
  • Change in velocity is defined as the difference between initial ($\vec{v_1}$) and final ($\vec{v_2}$) velocities or in math words as 

\[\Delta\vec{v}=\vec{v_2}-\vec{v_1}\]

  • 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. $\vec v=v$.  
  • The average acceleration vector points in the same direction as the vector $\Delta \vec{v}$.

By practicing these , you can master and understand more deeply. For a thorough and complete explanation related to the velocity and acceleration go to course pages. 

Some examples of average acceleration along the straight line:

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:
Acceleration is the change in velocity divided by the involved time. Since the car's velocity (magnitude and direction) is constant over the entire course, so by definition, its acceleration is zero. 

Example $2$: A plane has a take-off speed of $300\,{\rm \frac{km}{h}}$. What is the 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 ${\rm \frac{m}{s}}$ 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 elapsed time $\Delta t=45\,\rm s$ is defined as 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 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*}
That is 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 by dividing change of velocity, $\Delta v=20-10=10\,{\rm \frac ms}$ over the elapsed time $\Delta t=25\,{\rm s}$ 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 an acceleration of $1\,{\rm \frac{m}{s^2}}$?

Solution:
In this problem, the unknown quantity is the time interval over which acceleration occurs. Thus from definition of uniform 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 $\frac{10}{36}$.

Summary:

Velocity is a vector quantity with a magnitude and direction. If one of those changes during a finite period of time, then we have a new kinematic quantity which is known as average acceleration 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*}
In the case of infinitesimal time interval i.e. $\Delta t \to 0$, average acceleration is converted to instantaneous one by the following formula
 \[ a\equiv \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t}\]
SI unit: meter per second per second ($m/s^2$).
 



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