Equation Of Motions

Sir Isaac Newton's principles of motion became the foundation for all modern physical science, notably the study of mechanics. Newton's first law states that a body at rest or in motion will remain in that state unless an external force is applied. Thus, it is more than surprising that such laws underlying human physical motion dynamics can be mathematically solved by experts, allowing them to create systems that facilitate real-world applications and support our modern way of life.

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This Story also Contains

1. First Kinematical Equation Of Motion (Velocity-Time Equation)
2. Second kinematical Equation Of Motion (Position-Time Equation)
3. Third Kinematical Equation Of Motion (Velocity-Displacement Equation)
4. Displacement In The nth Second
5. Solved Example Based On Equation Of Motion
6. Summary

In this article, we will cover the concept of the equation of motion. This concept falls under the broader category of kinematics which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of 20 questions have been asked on this concept. And for NEET 4 questions were asked from this concept.

First Kinematical Equation Of Motion (Velocity-Time Equation)

Now first start with the first kinematics equation :

$v=u+at$

v = Final velocity

u = Initial velocity

a = acceleration

T = time

Let us understand the first equation of motion by the solved example given below:

Example : A particle has an initial velocity $3 \hat{i}+4 \hat{j}$ and an acceleration of $0.4 \hat{i}+0.3 \hat{j}$. Its speed after 10 s is
1) 10 units
2) $7 \sqrt{2}$ units
3) 7 units
4) 8.5 units

Solution:

Given :
$
\begin{aligned}
& \text { Initial velocity }(u)=3 \hat{i}+4 \hat{j} \\
& \text { Acceleration }(a)=0.4 \hat{i}+0.3 \hat{j} \\
& \text { Time }=10 \mathrm{~s} \\
& \text { using }, \vec{v}=\vec{u}+\vec{a} t \\
& \vec{v}=(3 i+4 j)+(0.4 i+0.3 j) \times 10 \\
& \vec{v}=(3 \hat{i}+4 \hat{j})+(4 \hat{i}+3 \hat{j}) \\
& \vec{v}=7 \hat{i}+7 \hat{j} \\
& |\vec{v}|=\sqrt{(7)^2+(7)^2}
\end{aligned}
$
so, speed is the magnitude of velocity-
$
|\vec{v}|=7 \sqrt{2} \text { unit }
$

Hence, the answer is the option (2).

Second kinematical Equation Of Motion (Position-Time Equation)

After reading the first equation of motion let's read the second equation of motion:

$ \begin{aligned}
& s=u t+\frac{1}{2} a t^2 \\
& s \rightarrow \text { Displacement } \\
& u \rightarrow \text { Initial velocity } \\
& a \rightarrow \text { acceleration } \\
& t \rightarrow \text { time }
\end{aligned} $

Let us understand the second equation of motion by the solved example:

Example 1: A particle starts from the origin at $\mathrm{t}=0$ with an initial velocity of $3.0 \hat{i} \mathrm{~m} / \mathrm{s}$ and moves in the $\mathrm{x}$-y plane with a constant acceleration $(6.0 \hat{i}+4.0 \hat{j}) \mathrm{m} / \mathrm{s}^2$. The $\boldsymbol{x}$-coordinate of the particle at the instant when its $y$-coordinate is $32 \mathrm{~m}$ is $D$ meters. The value of $D$ is :
1) 60
2) 50
3) 32
4) 40

Solution :

Given:
Taking motion along y -axis Displacement along y-axis $\left(s_y\right)=32 \mathrm{~m}$
Acceleration along y-axis, $\left(a_y\right)=4 \mathrm{~m} / \mathrm{s}^2$
Initial velocity along y-axis, $\left(u_y\right)=0 \mathrm{~m} / \mathrm{s}$
From 2nd equation of motion,
$
\begin{aligned}
& s_y=u_y t+\frac{1}{2} a_y t^2 \\
& \Rightarrow 32=0(t)+\frac{1}{2}(4)\left(t^2\right) \\
& \Rightarrow 32=2 t^2 \\
& \Rightarrow t=4 \mathrm{~s}
\end{aligned}
$

Now taking motion along $\mathrm{x}$ - axis.
Displacement along $\mathrm{x}$-axis $\left(\mathrm{s}_{\mathrm{x}}\right)=$ ?
Acceleration along x-axis , $\left(a_x\right)=6 \mathrm{~m} / \mathrm{s}^2$

initial velocity along $\mathrm{x}$-axis $\left(\mathrm{u}_{\mathrm{x}}\right)=3 \mathrm{~m} / \mathrm{s}$

So, $S_x=u_x t+\frac{1}{2} a_x t^2$ $=3 \times 4+\frac{1}{2} \times 16$ $=12+48$ $=60 \mathrm{~m}$

Hence, the answer is 60.

Example 2: Starting from the origin at time $t=0$, with initial velocity $5 \widehat{j} \mathrm{~ms}^{-1}$, a particle moves in the $\mathrm{x}-\mathrm{y}$ plane with constant acceleration of $(10 \hat{i}+4 \hat{j}) \mathrm{ms}^{-2}$. At time $\mathrm{t}$, its coordinates are $\left(20 \mathrm{~m}, y_0 \mathrm{~m}\right)$. The values of $\mathrm{t}$ and $y_0$ are, respectively:
1) $2 \mathrm{~s}$ and $18 \mathrm{~m}$
2) $4 \mathrm{~s}$ and $52 \mathrm{~m}$
3) $5 \mathrm{~s}$ and $24 \mathrm{~m}$
4) $5 \mathrm{~s}$ and $25 \mathrm{~m}$

Solution :

Given:
$
\begin{aligned}
\vec{a} & =(10 \hat{i}+4 \hat{j}) m s^{-2} \\
\vec{u} & =5 \hat{j} m s^{-1}
\end{aligned}
$

And final coordinates $\left(20, y_0\right)$ in time $t$
So
For $\mathrm{x}$-axis
$
\begin{aligned}
& \mathrm{S}_{\mathrm{x}}=u_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^2 \\
& 20-0=0+\left(\frac{1}{2} \times 10 \times \mathrm{t}^2\right) \\
& \mathrm{t}=2 \mathrm{sec}
\end{aligned}
$

For y-axis
$
\begin{aligned}
& \mathrm{S}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}} \times \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^2 \\
& \mathrm{y}_0=(5 \times 2)+\left(\frac{1}{2} \times 4 \times 2^2\right)=18 \mathrm{~m}
\end{aligned}
$

Hence, the answer is the Option 1.

Third Kinematical Equation Of Motion (Velocity-Displacement Equation)

Now we will study the the third equation of motion which is the Velocity-displacement equation:
$
v^2-u^2=2 a s
$
$v \rightarrow$ Final Velocity
$s \rightarrow$ Displacement
$u \rightarrow$ Initial velocity
$a \rightarrow$ acceleration

Displacement In The nth Second

We have now studied all three equations of motion. Now we'll look at the displacement in the nth second:

Formula: $S_n=u+\frac{a}{2}(2 n-1)$
Where $u=$ Initial velocity
$a=$ uniform acceleration

Recommended Topic Video

Solved Example Based On Equation Of Motion

Example 1: A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of the building simultaneously. The height of the building is :

1) 35 m

2) 45 m

3) 50 m

4) 25 m

Solution :

let the height of the tower be h

and the time taken to fall a distance of 5 m by the first stone is 1 sec

as

$
\begin{aligned}
& s=\frac{1}{2} g t_1^2 \\
& \Rightarrow 5=\frac{1}{2} * 10 * t_1^2 \\
& \Rightarrow t_1=1 \mathrm{sec}
\end{aligned}
$

Now, for the motion of the first stone
$
h=\frac{1}{2} g t^2 \ldots \ldots
$
and For the motion of the second stone
$
h-25=\frac{1}{2} g(t-1)^2 \text {. }
$

From equations (1) and (2)
we get $t=3$ sec
$
\text { So } h=\frac{1}{2} g t^2=5 *\left(3^2\right)=45 \mathrm{~m}
$

Hence, the answer is the Option (2).

Example 2: A ball is dropped from the top of a 100 m high tower on a planet. In the last seconds before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in ms^-2) near the surface of that planet is _______.

1) 8

2) 4

3) 2

4) 6

Solution :

Let the total time taken by the particle is t

So distance covered by a particle in $\mathrm{t} s e c=$
$
S_t=u t+\frac{1}{2} g t^2=\frac{1}{2} g t^2
$

Similarly, the distance covered by the particle in $\left(t-\frac{1}{2}\right) \mathrm{sec}=$ $S_{t-\frac{1}{2}}=\frac{1}{2} g\left(t-\frac{1}{2}\right)^2$

So distance covered by the particle in the last (1/2) sec of its journey $=S_t-S_{t-\frac{1}{2}}$

So,
$
19=S_t-S_{t-\frac{1}{2}}=\frac{1}{2} g t^2-\frac{1}{2} g\left(t-\frac{1}{2}\right)^2=\frac{1}{2} g\left(t-\frac{1}{4}\right) \ldots
$
and
$
S_t=\frac{1}{2} g t^2=100
$

From equations (1) and (2)
we get, $19 t^2-100 t+25=0 \Rightarrow t=5$, or $\quad t=0.26$ putting $\mathrm{t}=5$ we get $\mathrm{g}=8 \mathrm{~m} / \mathrm{s}^2$

Hence, the answer is 8.

Summary

The above Newton’s equations of motion are crucial for NEET and JEE exams, forming the foundation of the mechanics section of class 11. The first law addresses inertia, the second law relates force, mass, and acceleration, and the third law deals with action and reaction. Understanding these laws is essential for solving problems in physics, with applications ranging from simple motion to complex technical scenarios. Reading these concepts is vital for success in competitive exams and further studies in higher classes.