2.2: Accelerated Linear Motion and Generalization

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If you have understood the idea of taking the area under the velocity vs time graph, then this section would be quite simple to understand. And if you are familiar with basic calculus, it is easy to obtain a general solution for a velocity as the n^th degree function of time. However, for the sake of completeness and motivation, it will be covered here.

Accelerated Linear Motion

Accelerated linear motion means that the velocity itself is changing at a constant rate. This rate of change of velocity is called acceleration (denoted by \(\overrightarrow{a}\) ). We obtain the following velocity vs time graph for such motion,

The slope of the graph would be the acceleration (due to definition). To find the area under this curve, we employ 2 methods:

Method 1: Dividing into triangle and rectangle

From the graph, it is obvious that the area under this curve can be divided into ∆BCD and rectangle AOCD.

So, area of AOCD is OA×OC, i.e. \(\overrightarrow{u}\)×t ∵ \(\overrightarrow{u}\) is initial velocity

Area of ∆BCD is,

¹/₂×BD×CD

∴Area= ¹/₂×t×(\(\overrightarrow{a}\)t) ∵ Final velocity will be \(\overrightarrow{a}\)×t

Adding both the areas the final formula is,

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2\)

Remember this result.

Method 2: Integration

We know by the equation of a straight line that

\( \overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}t \) ∵ y=mx+c

From it, we obtain the value of the velocity at that instant of time. So, by integrating v with respect to time,

\(\int v \cdot dt = \int (\overrightarrow{u} + \overrightarrow{a}t) \cdot dt \)

We obtain the general solution,

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2 + s_0\) ∵ s₀is the constant of integration. For this equation, it represents the initial displacement

Generalizing the result

Using the equation for velocity frees us up from the need to draw a graph. Because of this, we can create a general equation that can give us the displacement for n^th degree time-dependent velocity. Let's take the example of increasing acceleration. The rate of change of acceleration is called jerk (j). So,

\( \overrightarrow{a} = \overrightarrow{a_0} + \overrightarrow{a}t \) ....(i)

\( \overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}t \) ....(ii)

By using (i) in (ii),

\(\overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}_0 t + \int \overrightarrow{j} \cdot dt \) ....(iii)

Plugging this into our previous formula for displacement and integrating we get,

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a_0}t^2 + \tfrac{1}{6} \overrightarrow{j}t^3 \)

From this, we can see an obvious pattern. Mathematically, it is expressed as,

\( \Sigma_{i=0}^{n} \frac{x_i}{i!}t^i \nonumber\)

Where 'x_i' represents the initial value of the i^th derivative of displacement. This formula can be used to solve any kinematics equation, but the math can turn quite hairy.

Or, if you are familiar with Taylor series, we can simply plug displacement into the equation to obtain:

\( \Sigma_{i=0}^{n} \frac{x_i}{i!}t^i \nonumber\)

This shows the power of knowing a little bit of calculus. Instead of integrating based on a graph, we were directly able to obtain the same definition of displacement.

Ending note

These derivations were all based on the simple formula, \(\overrightarrow{v} = \dfrac{\Delta \overrightarrow{s}}{\Delta \overrightarrow{t}}\)

With the help of some basic calculus, we were able to obtain the general equation for displacement. But the dependence of these equations on time can sometimes make the problem-solving process lengthy. So, we shall now derive a time-independent equation for uniformly accelerated linear motion.

\(\dfrac{\overrightarrow{v}-\overrightarrow{u}}{\overrightarrow{a}} = t \) ....(i)

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2\) ....(ii)

Using (i) in (ii) and simplifying, we obtain,

\(v^2 - u^2 = 2as \)

Notice that we haven't marked any of the values in the final formula as vectors. This is because for vectors, multiplication in the regular sense is meaningless (what does it mean to multiply two directions?). Therefore, we only use the magnitude of the vectors. More precisely, we use the dot product of the vectors. This shall be further discussed under vectors.

Though we have derived these formulae for only linear motion, because of vectors, they can be used in any scenario analyzing uniform acceleration.

Once again, the equations are:

i) \(\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t \)

ii) \(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2 + s_0\)

iii) \(v^2 - u^2 = 2as \)

The generalization of equation (i) and (iii) is similar to the generalization of (ii). With knowledge of integration, it is trivial and is left as an exercise for the reader.