2.10 Non-uniform acceleration

Problem : The acceleration of a particle along x-axis varies with time as :

$$a=t-1$$

Velocity and position both are zero at t= 0. Find displacement of the particle in the interval from t=1 to t = 3 s. Consider all measurements in SI unit.

Solution : Using integration result obtained earlier for expression of acceleration :

$$\text{Δ}v=v_2-v_1=\int a\left(t\right)đt$$

Here, $v_1=0$ . Let $v_2=v$ , then :

$$v=\underset{0}{\overset{t}{\int }}a\left(t\right)đt=\underset{0}{\overset{t}{\int }}\left(t-1\right)đt$$

Note that we have integrated RHS between time 0 ans t seconds in order to get an expression of velocity in “t”.

$$\Rightarrow v=\frac{t^2}{2}-t$$

Using integration result obtained earlier for expression of velocity : $$\text{Δ}x=\int v\left(t\right)đt=\int \left(\frac{t^2}{2}-t\right)dt$$

Integrating between t=1 and t=3, we have :

$$\text{Δ}x=\underset{1}{\overset{3}{\int }}\left(\frac{t^2}{2}-t\right)đt=[\frac{t^3}{6}-\frac{t^2}{2}\underset{1}{\overset{3}{]}}=\frac{27}{6}-\frac{9}{2}-\frac{1}{6}+\frac{1}{2}=5-4=1m$$

Problem : If the velocity of particle moving along a straight line is proportional to ${\frac{3}{4}}^{\mathrm{th}}$ power of time, then how do its displacement and acceleration vary with time?

Solution : Here, we need to find a higher order attribute (acceleration) and a lower order attribute displacement. We can find high order attribute by differentiation, whereas we can find lower order attribute by integration.

$$\begin{array}{l}v\propto t^{\frac{3}{4}}\end{array}$$

Let,

$$\begin{array}{l}v=k{t}^{\frac{3}{4}}\end{array}$$

where k is a constant. The acceleration of the particle is :

$$\begin{array}{l}a=\frac{đv\left(t\right)}{đt}=\frac{3k{t}^{\frac{3}{4}-1}}{4}=\frac{3k{t}^{-\frac{1}{4}}}{4}\end{array}$$

Hence,

$$\begin{array}{l}v\propto t^{-\frac{1}{4}}\end{array}$$

For lower order attribute “displacement”, we integrate the function :

$$\begin{array}{l}đx=vđt\\ \Rightarrow \Delta x=\int vđt=\int k{t}^{\frac{3}{4}}đt=\frac{4k{t}^{\frac{7}{4}}}{7}\\ \Delta x\propto t^{\frac{7}{4}}\end{array}$$

Problem : The acceleration of a particle along x-axis varies with position as :

$$a=9x$$

Velocity is zero at t = 0 and x=2m. Find speed at x=4. Consider all measurements in SI unit.

Solution : Using integration,

$$\int vđv=\int a\left(x\right)đx$$ $$\underset{0}{\overset{v}{\int }}vđv=\underset{2}{\overset{x}{\int }}9xđx$$ $$\Rightarrow [\frac{v^2}{2}\underset{0}{\overset{v}{]}}=9[\frac{x^2}{2}\underset{2}{\overset{x}{]}}$$ $$\Rightarrow \frac{v^2}{2}=\frac{9x^2}{2}-\frac{36}{2}$$ $$\Rightarrow v^2=9x^2-36=9\left(x^2-4\right)$$ $$\Rightarrow v=\pm 3\sqrt{\left(x^2-4\right)}$$

We neglect negative sign as particle is moving in x-direction with positive acceleration. At x =4 m,

$$\Rightarrow v=3\sqrt{\left(4^2-4\right)}=3\sqrt{12}=6\sqrt{3}m/s$$