8.1 Solved examples

  Motion in One Dimension

    1.0 Introduction
    2.0 Kinematic variables
    3.0 Motion in one dimension
    4.0 Derivation of the kinematics equation
    5.0 Vertical motion under gravity
    6.0 Analysis of motion through graph
    7.0 Relative motion
    8.0 Simultaneous motion of two bodies
    9.0 River boat problem
    10.0 Aircraft-wind problem
    11.0 Rain problem

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8.1 Solved examples

Question: A police jeep is chasing a thief going on a motorbike. The thief crosses a turning at a speed of $20\ m/s$. The police follow it at a speed of $25\ m/s$, crossing the turning ten seconds later than the thief. Assuming that they travel at constant speeds. Find the time and how far from the turning will the police catch up the thief.

Solution:

We can write all the information as below,

${\overrightarrow d _{TP}} = 200\widehat i$
${\overrightarrow v _P} = 25\widehat i$
${\overrightarrow v _T} = 20\widehat i$

Let us make police as the observer. Velocity of thief with respect to police is,
$${\overrightarrow v _{TP}} = {\overrightarrow v _T} - {\overrightarrow v _P}$$$${\overrightarrow v _{TP}} = 20\widehat i - 25\widehat i$$$${\overrightarrow v _{TP}} = - 5\widehat i$$

Time after which the police will catch the thief is,
$$t = \frac{{200}}{5} = 40s$$
So, the distance traveled by the police in this time interval will be,
$${d_P} = {v_P}t$$$${d_P} = 25 \times 40$$$${d_P} = 1000m$$or$${d_P} = 1km$$

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Question: The elevator of height $8\ m$ starts to move downward with an acceleration of $9\ m/s^2$. At the same time, lamp is detached from the ceiling of the elevator. Find the time after which the lamp will strike the floor of the elevator. $\left( {g = 10m/{s^2}} \right)$

Solution:

Consider the lamp and the elevator as two separate bodies.

Acceleration of the lamp wrt elevator is,
$${\overrightarrow a _{LE}} = {\overrightarrow a _L} - {\overrightarrow a _E}$$$${\overrightarrow a _{LE}} = - 10\widehat j - \left( { - 9} \right)\widehat j$$$${\overrightarrow a _{LE}} = - \widehat j$$

We can write all the information as below,

${\overrightarrow u _{FL}} = 0$
${\overrightarrow a _{FL}} = - \widehat j$
${\overrightarrow s _{FL}} = - 8\widehat j$
$t = ?$

Kinematic equation relating displacement and time as,
$${\overrightarrow s _{LE}} = {\overrightarrow u _{LE}}t + \frac{1}{2}{\overrightarrow a _{LE}}{t^2}$$$$ - 8 = 0 + \frac{1}{2}\left( { - 1} \right){t^2}$$$${t^2} = 16$$$$t = 4s$$

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Question: The distance between two moving particles at any time is $a$. If $v$ be their initial velocity and $v_1$ and $v_2$ be the component of $v$ along and perpendicular to $a$. Find the time when they are closest to each other. Also find the minimum distance between them.

Solution: Let the two particles be $A$ and $B$. Consider the relative velocities is wrt to $B$.

Now we can understand the situation better with the diagram.

We can write, $$v = \sqrt {v_1^2 + v_2^2} $$

After anytime $t$,

As observed by $B$, the particle $A$ is moving along $AC$ i.e. blue line.

The perpendicular from $B$ on the path will give us the shortest distance $(r)$.

The particle has traveled distance $\left( {{v_1}t} \right)$ along $x-$ axis and $\left( {{v_2}t} \right)$ along $y-$ axis.

From the figure we can write,
$$r = \sqrt {{{\left( {a - {v_1}t} \right)}^2} + {{\left( {{v_2}t} \right)}^2}} $$
For finding minimum value of $r$, we will differentiate it wrt $t$.
$$\frac{{dr}}{{dt}} = \frac{{2\left( {a - {v_1}t} \right)\left( { - {v_1}} \right) + 2v_2^2t}}{{2\sqrt {{{\left( {a - {v_1}t} \right)}^2} + {{\left( {{v_2}t} \right)}^2}} }}$$$$0 = \frac{{2\left[ { - a{v_1} + v_1^2t + v_2^2t} \right]}}{{2\sqrt {{{\left( {a - {v_1}t} \right)}^2} + {{\left( {{v_2}t} \right)}^2}} }}$$$$\left( {v_1^2 + v_2^2} \right)t = a{v_1}$$
The time at which the minimum distance is,
$$t = \frac{{a{v_1}}}{{{v^2}}}$$
Value of the minimum distance is,
$$r = \sqrt {{{\left( {a - {v_1}t} \right)}^2} + {{\left( {{v_2}t} \right)}^2}} $$$$r = \sqrt {{{\left( {a - {v_1} \times \frac{{a{v_1}}}{{{v^2}}}} \right)}^2} + {{\left( {{v_2} \times \frac{{a{v_1}}}{{{v^2}}}} \right)}^2}} $$$$r = \sqrt {{{\left( {\frac{{a{v^2} - av_1^2}}{{{v^2}}}} \right)}^2} + {{\left( {\frac{{a{v_1}{v_2}}}{{{v^2}}}} \right)}^2}} $$$$r = a\sqrt {\left( {\frac{{{v^4} + v_1^4 - 2v_1^2{v^2} + v_1^2v_2^2}}{{{v^4}}}} \right)} $$$$r = a\sqrt {\left( {\frac{{{{\left( {{v^2}} \right)}^2} + v_1^4 - 2v_1^2{v^2} + v_1^2v_2^2}}{{{v^4}}}} \right)} $$As $\left( {{v^2} = v_1^2 + v_2^2} \right)$, $$r = a\sqrt {\left( {\frac{{{{\left( {v_1^2 + v_2^2} \right)}^2} + v_1^4 - 2v_1^2\left( {v_1^2 + v_2^2} \right) + v_1^2v_2^2}}{{{v^4}}}} \right)} $$$$r = a\sqrt {\left( {\frac{{v_1^4 + v_2^4 + 2v_1^2v_2^2 + v_1^4 - 2v_1^4 - 2v_1^2v_2^2 + v_1^2v_2^2}}{{{v^4}}}} \right)} $$$$r = a\sqrt {\left( {\frac{{2v_1^4 + v_2^4 + 2v_1^2v_2^2 - 2v_1^4 - 2v_1^2v_2^2 + v_1^2v_2^2}}{{{v^4}}}} \right)} $$$$r = a\sqrt {\left( {\frac{{v_2^4 + v_1^2v_2^2}}{{{v^4}}}} \right)} $$$$r = a\sqrt {\frac{{v_2^2\left( {v_2^2 + v_1^2} \right)}}{{{v^4}}}} $$$$r = a\sqrt {\frac{{v_2^2{v^2}}}{{{v^4}}}} $$$$r = a\sqrt {\frac{{v_2^2}}{{{v^2}}}} $$$$r = \frac{{a{v_2}}}{v}$$

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Question: Two cars $P$ and $Q$ start moving simultaneously in the same direction with acceleration of $1\,m/{s^2}$ & $2\,m/{s^2}$ and $3\,m/s$ & $1\,m/s$ respectively. Initially $P$ is $10\ m$ behind $Q$. What is the minimum distance between them?

Solution:

We can write all the information below,

${\overrightarrow u _P} = 3\widehat i$
${\overrightarrow u _Q} = 1\widehat i$

Velocity of car $P$ wrt $Q$,
$${\overrightarrow u _{PQ}} = {\overrightarrow u _P} - {\overrightarrow u _Q}$$$${\overrightarrow u _{PQ}} = 3\widehat i - \widehat i$$$${\overrightarrow u _{PQ}} = 2\widehat i \quad ...(i)$$

Similarly,

${\overrightarrow a _P} = \widehat i$
${\overrightarrow a _Q} = 2\widehat i$

Acceleration of car $P$ wrt $Q$,
$${\overrightarrow a _{PQ}} = {\overrightarrow a _P} - {\overrightarrow a _Q}$$$${\overrightarrow a _{PQ}} = \widehat i - 2\widehat i$$$${\overrightarrow a _{PQ}} = -\widehat i \quad ...(ii)$$

Till the time car $P$ move towards $Q$, the distance between them decreases. When the relative velocity between the cars becomes zero, they will have minimum separation after that the distance between them starts increasing.

So, we can write all the information as below.

${\overrightarrow u _{PQ}} = 2\widehat i$
${\overrightarrow a _{PQ}} = -\widehat i$
${\overrightarrow v _{PQ}} = 0$

Let the displacement be $s$. So, we can write,
$$v_{PQ}^2 = u_{PQ}^2 + 2{\overrightarrow a _{PQ}}{\overrightarrow s _{PQ}}$$$$0 = {\left( 2 \right)^2} + 2\left( { - 1} \right)s$$$$s = 2\,m$$
So, the minimum distance between them is,
$$d=(10-2)\ m$$$$d=8\ m$$