Whenever we consider the motion of a moving object, it is rare that the velocity will remain constant throughout its motion. The speed of objects typically increases and decreases over the course of their trajectories. Acceleration is the word used to refer to the rate of change of speed and it is a measure of the rate at which the speed of an object is increasing or decreasing. This is called acceleration. It is used in a lot of important calculations like when designing the braking system of a vehicle etc. In this article, we will look into the different equations that are used in calculating the acceleration of a body. We will also go through a few real-life examples where use the equations.
- Acceleration definition
- Acceleration vector
- Velocity and acceleration time graphs
- Acceleration formula
- Acceleration due to Gravity
Acceleration definition
Acceleration is the rate of change of velocity with respect to time
We can calculate the acceleration if we know how much an object's velocity changes over a period of time given that it is moving in a straight line with a constant acceleration. It is given by the following equation
\[a=\dfrac{v-u}{t}\]
or in words,
\[\text{Acceleration}=\dfrac{\text{Change in velocity}}{\text{Time taken}}\]
where \(v\) is the final velocity, \(u\) is the initial velocity of the object and \(t\) is the time taken for the object to change in velocity from \(u\) to \(v\).
Acceleration Units
The SI units of acceleration are \(\mathrm{m}/\mathrm{s}^2\). Acceleration can be negative or positive. Negative acceleration is called deceleration.
Acceleration vector
Acceleration \(\vec{a}\) is a vector quantity. This is also because it is derived from the velocity vector \(\vec{v}\). Looking at the equation for the acceleration vector we can see that it is directly proportional to the change of velocity and inversely proportional to the time it takes to accelerate or decelerate. In fact, we can get a sense of the direction of the acceleration vector by looking at the magnitude of the velocity vector.
If the velocity of an object is increasing (initial velocity < final velocity) then it has a positive acceleration in the direction of velocity.
If the velocity is decreasing, (\(u>v\)) then the acceleration is negative and in the opposite direction of velocity.
If the velocity is uniform (\(u=v\)) then the acceleration is \(0\). Why do you think so? This is because acceleration is given by the change in velocity. Let us visualize this relation using graphs.
\[a=\dfrac{v-u}{t},\quad\text{if}\quad v-u=0,\quad\text{then}\quad a=0\]
Velocity and acceleration time graphs
The Velocity and acceleration of a moving object can be visualized using a time graph. The graph below shows the velocity-time graph of an object moving in a straight line.
Velocity-time graph with three sections corresponding to acceleration, constant velocity and deceleration, Kids Brittanica
The orange line indicates that the velocity is increasing with respect to time this means that the object has positive acceleration.
The green line is parallel meaning that the velocity is constant which means that the acceleration is Zero.
The blue line is a downward slope that shows the velocity decreasing this is indicative of negative deceleration.
To calculate the acceleration at any point we need to find the slope of the velocity curve.
\[\text{slope}=\dfrac{y_2-y_1}{x_2-x_1}\]
where \((x_1,y_1)\) are the coordinates of the initial point on the graph and \((x_2,y_2)\) are the coordinates of the final point. We know that the y-axis records velocity and the x-axis record the time taken, this means that the formula is nothing but:
\[a=\dfrac{v-u}{t}\]
Let us look at this as an example.
Find the acceleration of the object from the above velocity-time graph for the initial \(10\) seconds.
Solution
The acceleration between two points = slope of the velocity-time graph. The formula for the slope of the velocity-time graph is given by
\[\begin{align} a(\text{slope})&=\dfrac{y_2-y_1}{x_2-x_1}=\\&=\dfrac{5-0}{10-0}=\\&=0.5\,\mathrm{m/s}^2\end{align}\]
The acceleration time graph gives the acceleration of the body with respect to time. We can also calculate the velocity by estimating the slope of the graph, StudySmarter Originals
We can see the acceleration is constant for the first \(5\,\mathrm{s}\) as the object increases its velocity from \(0\) to \(5\, \mathrm{m/s}\). Next, there is a sudden drop to zero for a period of \(10\,\mathrm{s}\) when the velocity is constant and finally, the acceleration drops to \(-0.5\,\mathrm{m/s}^2\) when the object decelerates from \(5\,\mathrm{m/s}\) to \(10\,\mathrm{m/s}\). To calculate the velocity at any point all you have to do is find the area under the acceleration curve. Let us now work on a few examples using the above equations.
A car accelerates in a time of \(10\,\mathrm{s}\) from \(10\,\mathrm{m/s}\) to \(15\,\mathrm{m/s}\). What is the acceleration of the car?
Step 1: Write down the given quantities
\[v=15\,\tfrac{\mathrm{m}}{\mathrm{s}},\quad u=10\tfrac{\mathrm{m}}{\mathrm{s}},\quad t=10\, \mathrm{s}\]
Now using the equation for acceleration,
\[\begin{align}a&=\dfrac{v-u}{t}=\\&=\dfrac{15\,\mathrm{m}/\mathrm{s}-10\,\mathrm{m}/\mathrm{s}}{10\,\mathrm{m}/\mathrm{s}}=\\&=\dfrac{5\,\mathrm{m}/\mathrm{s}}{10\,\mathrm{s}}=0.5\,\mathrm{s}/\mathrm{s}^2\end{align}\]
To put this into perspective, the acceleration due to gravity (\(g\)) is \(9.8\,\mathrm{m}/\mathrm{s}^2\). Which makes the acceleration of the car approximately \(0.05g\), where \(g\) is the acceleration is due to gravity at the surface of the Earth \((\approx 9.81\,\mathrm{m}/\mathrm{s}^2)\).
Acceleration formula
Now we know some of the relations between acceleration, velocity, and time. But is it possible to relate distance travelled directly with acceleration? Assume an object starts from rest (initial velocity, \(u=0\)) and then accelerates to a final velocity \(v\) in time \(t\). The average velocity is given by
\[v_{\text{average}}=\dfrac{s}{t}\]
Re-arranging the equation for the distance \(s\) we get
\[s=v_{\text{average}}t\]
The acceleration of the object is equal to \(\dfrac{v-0}{t}\) as it started from rest \((u=0)\).
\[a=\dfrac{v}{t}\]
Re-arranging in terms of \(v\) we get
\[v=at\]
The average velocity of the object is given by
\[v_{\text{average}}=\dfrac{v+u}{2}=\dfrac{v_f}{2}\]
Plug the average velocity in the above equation and we get
\[v_{\text{average}}=2at\]
Finally, plug this in the equation for the distance and we obtain
\[s=\dfrac{1}{2}at^2\]
There you have it, an equation that directly relates acceleration and displacement. But what if the object did not start moving from rest? i.e. \(v_i\) is not equal to \(0\). Let’s work it out. The acceleration is now equal to
\[a=\dfrac{v-u}{t}\]
Rearrange for final velocity \(v\), and we obtain,
\[v=u+at\]
The average velocity changes to
\[a_{\text{average}}=\dfrac{u+v}{2}\]
Plug the value for final velocity in the above equation
\[v_{\text{average}}=\dfrac{u+u+at}{2}=u+\dfrac{1}{2}at\]
The equation for distance travelled is still
\[s=v_{\text{average}}t\]
Plug the equation for \(v_{\text{average}}\) in the formula for distance and we obtain
\[s=\left(u+\dfrac{1}{2}at\right)t\]
\[s=ut+\dfrac{1}{2}at^2\]
The above equation relates to distance and acceleration when an object already has some initial velocity. That’s it if you look at it from another angle ut is just the distance during initial velocity. Add this to the distance travelled during final velocity \(\frac{1}{2}at^2\). Unfortunately, we have one last equation this equation relates to acceleration distance and velocity altogether. How interesting is that? Here’s how it works; first, you rearrange the equation for acceleration with respect to the time:
\[t=\dfrac{v-u}{a}\]
Now displacement,
\[s=v_{\text{average}}t\]
And the average velocity when acceleration is constant is given by
\[v_{\text{average}}=\dfrac{1}{2}(v+u)\]
Substitute \(V_{\text{average}}\) in the equation for \(s\) and we get
\[s=\dfrac{1}{2}(v+u)t\]
Substituting for the time, you get
\[s=\dfrac{1}{2}(v+u)t\]
\[s=\dfrac{1}{2}\dfrac{(v+u)(v-u)}{a}\]
Simplifying using the laws of algebra, we get
\[s=\dfrac{1}{2}\dfrac{v^2-u^2}{a}\]
\[2as=v^2-u^2\]
There, you have three new equations that you can use to find acceleration velocity and distance. Understanding how these equations work as compared to trying to memorize them gives you more control and flexibility while solving problems. Now let us look at an example that will test your understanding of when to use the right formula,
A car begins at a speed of \(3\,\mathrm{m}/\mathrm{s}\) and accelerates at \(2\,\mathrm{s}/\mathrm{s}^2\) over a distance of\(40\,\mathrm{m}\),
calculate the final speed of the car.
Step 1: Write down the given quantities
\[u=3\,\mathrm{m}/\mathrm{s},\quad a=2\,\mathrm{m}/\mathrm{s}^2,\quad s=40\,\mathrm{m},\quad v=?\]
Step 2: Use the appropriate equation for calculating the final velocity of the car
In the above problem, we have the values of initial velocity, acceleration and time hence we can use the following equation to find the final velocity
\[\begin{align} v^2-u^2&=2as\\v&=\sqrt{\dfrac{2as}{u^2}}\\v&=\sqrt{\dfrac{2\times 2\,\mathrm{m}/\mathrm{s}^2\times 40\,\mathrm{m}}{3\,\mathrm{s}/\mathrm{s}\times 3\,\mathrm{m}/\mathrm{s}}}\\v&=4.21\,\mathrm{m}/\mathrm{s}\end{align}\]
The final velocity of the car is \(4.21\,\mathrm{m}/\mathrm{s}\).
Acceleration due to Gravity
The acceleration due to gravity represented by \(g\) is the acceleration of an object when it's free-falling due to the gravitational force acting on it. This acceleration due to gravity depends on the gravitational force exerted by the planet. Hence it will change for different planets. The standard value of \(g\) on earth is considered to be \(9.8\,\mathrm{m}/\mathrm{s}^2\). What does that mean? This implies that a free-falling object will accelerate at the value of \(g\) as it keeps falling towards the earth.
The value of \(g\) as we know is constant, but it actually changes due to a lot of factors. The value of \(g\) is affected by depth or altitude. The value of \(g\) decreases as the depth of the object increases. It can also be affected by its position on the Earth. The value of \(g\) is more on the equator than on the poles. And finally, this value is also affected due to the rotation of the earth.
This brings us to the end of this article let’s look at what we’ve learned so far.
Acceleration - Key takeaways
- Acceleration is the rate of change of velocity with respect to time.
- Acceleration is given by \(a=\dfrac{v-u}{t}\) and is measured in \(\mathrm{m}/\mathrm{s}^2\).
- The velocity and acceleration of a moving object can be visualized using an acceleration-time graph.
- To calculate the acceleration at any point we need to find the slope of the velocity-time curve using the equation \(a(\text{slope})=\dfrac{v_1-v_2}{t_1-t_2}\).
- To calculate the velocity from the acceleration-time graph we calculate the area under the acceleration curve.
- The relationship between acceleration, distance and velocity is given by the following equations \(s=\dfrac{1}{2}at^2\) ( when the object starts from rest) and \(s=ut+\dfrac{1}{2}at^2\)(when the object is in motion) and \(2as=v^2-u^2\).
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