2.6 One dimensional motion with constant acceleration  (Page 4/6)

Problem : The equation of motion for displacement in $n^{\mathrm{th}}$ second is given by :

$$\begin{array}{l}{x}_n=u+\frac{a}{2}(2n-1)\end{array}$$

This equation is dimensionally incompatible, yet correct. Explain.

Solution : Since “n” is a number, the dimension of the terms of the equation is indicated as :

$$\begin{array}{l}\left[x_n\right]=\left[\mathrm{Displacment}\right]=\left[L\right]\\ \left[u\right]=\left[\mathrm{velocity}\right]=\left[L{T}^{-1}\right]\\ \left[\frac{a}{2}\right(2n-1\left)\right]=\left[\mathrm{acceleration}\right]=\left[L{T}^{-2}\right]\end{array}$$

Clearly, dimensions of the terms are not same and hence equation is apparently incompatible in terms of dimensions.

In order to resolve this apparent incompatibility, we need to show that each term of right hand side of the equation has dimension that of displacement i.e. length. Now, we know that the relation is derived for a displacement for time equal to “1” second. As multiplication of “1” or " $1^2$ " with any term is not visible, the apparent discrepancy has appeared in otherwise correct equation. We can, therefore, re-write the equation with explicit time as :

$$\begin{array}{l}{x}_n=ux1+\frac{a}{2}(2n-1)x{1}^2\\ \Rightarrow x_n=uxt+\frac{a}{2}(2n-1)x{t}^2\end{array}$$

Now, the dimensions of the first and second terms on the right side are :

$$\begin{array}{l}[uxt]=\left[L{T}^{-1}T\right]=\left[L\right]\\ \left[\frac{a}{2}\right(2n-1)x{t}^2]=\left[L{T}^{-2}{T}^2\right]=\left[L\right]\end{array}$$

Thus, we see that the equation is implicitly correct in terms of dimensions.

Problem : A particle starting with velocity “u” covers two equal distances in a straight line with accelerations $a_1$ and $a_2$ . What is the equivalent acceleration for the complete motion?

Solution : The equivalent acceleration for the complete motion should yield same value for the attributes of the motion at the end of the journey. For example, the final velocity at the end of the journey with the equivalent acceleration should be same as the one calculated with individual accelerations.

Here initial velocity is "u". The velocity at the end of half of the distance (say “x”) is :

$$\begin{array}{l}{v_1}^2=u^2+2a_{1}x\end{array}$$

Clearly, $v_1$ is the initial velocity for the second leg of motion,

$$\begin{array}{l}{v_2}^2={v_1}^2+2a_{2}x\end{array}$$

Adding above two equations, we have :

$$\begin{array}{l}{v_2}^2=u^2+2(a_1+a_2)x\end{array}$$

This is the square of velocity at the end of journey. Now, let “a” be the equivalent acceleration, then applying equation of motion for the whole distance (2x),

$$\begin{array}{l}{v_2}^2=u^2+2a\left(2x\right)\end{array}$$

Comparing equations,

$$\begin{array}{l}a=\frac{a_1+a_2}{2}\end{array}$$