Acceleration

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.

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.

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,