6.4 Solved examples

  Motion in One Dimension

Question: The velocity-time graph of a linear motion is as shown in the figure. Find the distance traveled and displacement of the particle from the origin after $8\ s$.

Solution:

The vector sum of area under the velocity-time graph gives displacement $(d)$.
$$d = ar\left( {OABC} \right) + ar\left( {CDEF} \right)$$$$d = \left[ {\left( {\frac{1}{2} \times 1 \times 4} \right) + \left( {2 \times 4} \right) + \left( {\frac{1}{2} \times 1 \times 4} \right)} \right] + \left[ {\left( {\frac{1}{2} \times 1 \times - 2} \right) + \left( {2 \times - 3} \right) + \left( {\frac{1}{2} \times 1 \times - 2} \right)} \right]$$$$d = \left[ {2 + 8 + 2} \right] + \left[ { - 1 - 6 - 1} \right]$$$$d = 12 - 8$$$$d = 4\,m$$

The modulus sum of area under the graph gives us distance $(D)$.

$$d = \left| {ar\left( {OABC} \right)} \right| + \left| {ar\left( {CDEF} \right)} \right|$$$$d = \left[ {\left( {\frac{1}{2} \times 1 \times 4} \right) + \left( {2 \times 4} \right) + \left( {\frac{1}{2} \times 1 \times 4} \right)} \right] + \left| {\left[ {\left( {\frac{1}{2} \times 1 \times - 2} \right) + \left( {2 \times - 3} \right) + \left( {\frac{1}{2} \times 1 \times - 2} \right)} \right]} \right|$$$$d = \left[ {2 + 8 + 2} \right] + \left| {\left[ { - 1 - 6 - 1} \right]} \right|$$$$d = 12 + 8$$$$d = 20\,m$$

Question: The acceleration displacement graph of a particle moving in a straight line is shown in the figure. Initial velocity of the particle is zero. Find the velocity of the particle when displacement of the particle is $S=12\ m$.

Solution:

The area under the graph will give us product of acceleration and displacement i.e. $\overrightarrow a .\,\overrightarrow s $

So, $$\overrightarrow a .\,\overrightarrow s = {A_1} + {A_2} + {A_3} + {A_4} + {A_5}$$

${A_1} = \frac{1}{2} \times 2 \times 2 = 2\,units$
${A_2} = 2 \times 6 = 12\,units$
${A_3} = 2 \times 2 = 4\,units$
${A_4} = \frac{1}{2} \times 2 \times 2 = 2\,units$
${A_5} = \frac{1}{2} \times 2 \times 4 = 4\,units$

Therefore, $$\overrightarrow a .\,\overrightarrow s = 2 + 12 + 4 + 2 + 4$$$$\overrightarrow a .\,\overrightarrow s = 24\,units$$

We can write all the information below,

$\overrightarrow u = 0$
$\overrightarrow v = ?$
$\overrightarrow a .\,\overrightarrow s = 24\,units$

We can write the kinematic equation as,
$${v^2} = {u^2} + 2\overrightarrow a .\,\overrightarrow s $$$${v^2} = 0 + 2 \times 24$$$${v^2} = 48$$$$v = 4\sqrt 3 m/s$$

Question: If the velocity $v$ of a particle moving along a straight line decreases lineraly with its displacement $s$ as shown in the figure. Find the acceleration of the particle at $s=15\ m$.

Solution: The equation of the slope is a straight line. So, it should be in the form of $y=mx+c$

Slope is given by,
$$m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{10 - 0}}{{0 - 30}} = - \frac{1}{3}$$
$y$ intercept is,
$$c = 10$$
So, the equation becomes, $$y = - \frac{x}{3} + 10$$or$$v = - \frac{s}{3} + 10$$
When displacement is $s=15\ m$, then velocity is $v=5\ m/s$

Also, can be written as,
$$m = \frac{{dv}}{{ds}}$$$$ - \frac{1}{3} = \frac{{dv}}{{dt}} \times \frac{{dt}}{{ds}}$$$$ - \frac{1}{3} = a\left( {\frac{1}{v}} \right)$$$$a = - \frac{v}{3}$$So, acceleration at $s=15\ m$ is,
$$a = - \frac{5}{3}\,m/{s^2}$$