3.2 Motion in a straight line with uniform acceleration

  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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3.2 Motion in a straight line with uniform acceleration

Kinematics equation for a body undergoing uniform acceleration is given by,
$$\overrightarrow v = \overrightarrow u + \overrightarrow a t$$$$\overrightarrow s = \overrightarrow u t + \frac{1}{2}\overrightarrow a {t^2}$$$${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $$$${\vec s_{{n^{th}}}} = \vec u + \frac{1}{2}\vec a\left( {2n - 1} \right)$$
where,
$\overrightarrow u :$ Initial velcoity of a particle
$\overrightarrow v :$ Final velcoity of a particle
$\overrightarrow a :$ Constant acceleration
$t:$ Any time $t$
${\overrightarrow S _{n^{th}}}:$ Displacement of a particle at $n^{th}$ second
$\overrightarrow S :$ Displacement of a particle in time $t$

Note:

• All the above equations are valid only when acceleration is constant.
• $\overrightarrow v $, $\overrightarrow u $ and $\overrightarrow a $ are vector quantities

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Question: The speed of the train is reduced from $10\ m/s$ to $2.5 \ m/s$ whilst it travels a distance of $150\ m$, if the retardation to be uniform. Find how much further it will travel before coming to rest.

Solution: We can write all the information as,

$u = 10\,m/s$
$v = 2.5\,m/s$
$s=150\ m$
$a=?$

So, writing the kinematics equation relating displacement and acceleration as below,
$${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $$$${\left( {2.5} \right)^2} = {\left( {10} \right)^2} + 2a \times 150$$$$6.25 = 100 + 300a$$$$300a = - 93.75$$$$a = - \frac{{93.75}}{{300}}m/{s^2}$$$$a = - 0.3125m/{s^2}$$

Now when the particle is coming to rest.

$u = 2.5\,m/s$
$v = 0\,m/s$
$a=-0.3125\ m/s^2$
$s=?$

So we can write,
$${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $$$${0^2} = {\left( {2.5} \right)^2} + 2 \times \left( { - 0.3125} \right) \times s$$$$ - 6.25 = - 0.625s$$$$s = \frac{{ - 6.25}}{{ - 0.625}}$$$$s = 10m$$

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Question: A train stops at two stations $P$ and $Q$ which are $2\ km$ apart. It accelerates uniformly from $P$ at $1\ m/s^2$ for $15\ s$ and maintains a constant speed for a time before decelerating uniformly to rest at $Q$. If the deceleration is $0.5\ m/s^2$, find the time for which the train is travelling at a constant speed.

Solution: We can visualize the above question in the figure as,

From the figure we can write,
$${d_1} + {d_2} + {d_3} = 2000m\quad ...(i)$$
From $P$ to $R$

As the train starts from rest. We can write the following information.

$u_P=0$
$v_R=?$
$a_1=1\ m/s^2$
$t_1=15\ s$
$d_1=?$

We can write velocity-time equation as,
$$v = u + at$$$$v_R = 0 + 1 \times 15$$$$v_R = 15m/s$$
Similarly displacement-time equation can be written as,
$$s = ut + \frac{1}{2}a{t^2}$$$${d_1} = 0 + \frac{1}{2} \times 1 \times {15^2}$$$${d_1} = 112.5m \quad ...(ii)$$
From $R$ to $S$

Between point $R$ and $S$, the train is travelling with constant velocity. Let the time be $t_2$. So,
$${d_2} = 15{t_2} \quad ...(iii)$$
From $S$ to $Q$

Now the train starts decelerating and comes to rest at $Q$. We can write the following information as,

$u_S=15\ m/s$
$v_Q=0$
$a_3=-0.5\ m/s^2$
$d_3=?$

We can write the velocity-displacement relations as,
$${v^2} = {u^2} + 2as$$$$0 = {\left( {15} \right)^2} + 2\left( { - 0.5} \right){d_3}$$$${d_3} = 225m\quad ...(iv)$$
From equation $(i)$, $(ii)$, $(iii)$ and $(iv)$ we can write,
$$112.5 + 15{t_2} + 225 = 2000$$$$15{t_2} = 2000 - \left( {112.5 + 225} \right)$$$$15{t_2} = 1662.5$$$${t_2} = 110.83s$$

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Question: A particle is moving with uniform acceleration, in eighth and thirteenth second after starting it moves through $255\ cm$ and $225\ cm$ respectively. Find its acceleration $(cm/s^2)$.

Solution: Let the initial velocity and acceleration be $u$ and $a$ respectively.

We can write the equation for displacement in $n^{th}$ second as,
$${s_{{n^{th}}}} = u + \frac{1}{2}a\left( {2n - 1} \right)$$ So, $${s_{{8^{th}}}} = u + \frac{1}{2}a\left( {2 \times 8 - 1} \right)$$or$$255 = u + \frac{1}{2}a\left( {2 \times 8 - 1} \right)\quad ...(i)$$Similarly,$${s_{{{13}^{th}}}} = u + \frac{1}{2}a\left( {2 \times 13 - 1} \right)$$or$$225 = u + \frac{1}{2}a\left( {2 \times 13 - 1} \right)\quad ...(ii)$$ Subtracting equation $(i)$ and $(ii)$ we get,
$$255 - 225 = \left[ {u + \frac{1}{2}a\left( {2 \times 8 - 1} \right)} \right] - \left[ {u + \frac{1}{2}a\left( {2 \times 13 - 1} \right)} \right]$$$$30 = \frac{1}{2}a\left( {16 - 26} \right)$$$$30 = \frac{1}{2}a \times \left( { - 10} \right)$$$$a = - 6cm/{s^2}$$