Table of Contents
Introduction

In the previous section, we defined the kinematic quantities used in the study of kinematics, including displacement, velocity and acceleration and the relationships between these quantities.

In this section, we derive the kinematic equations of motion and demonstrate their applications for two special cases:
 uniform motion
 uniformly accelerated motion.
Uniform Motion
 By uniform motion, we mean a motion where the velocity is constant. In other words, the $x$$t$ graph is a straight line with a constant slope. In equal intervals of time $\Delta t$, the position changes by the same amount $\Delta x$.
Equations of Motion
 We recall the definition of velocity,
$$ v(t)=\frac{dx(t)}{dt}. $$

We now express $dx(t)$ as the subject and integrate both sides:
$$ \int_{x_0}^{x_f} dx(t) = \int_{t_i}^{t_f} v(t)dt. $$
where $x_0$ and $x_f$ are the initial and final positions corresponding to the initial and final times $t_i$ and $t_f$, respectively.

Since we are dealing with uniform motion, $v(t) = \overline{v}$ is a constant and can be taken out of the integral. Furthermore, we start our clock when motion starts: $t_i=0$. Finally, we let the final time and position be $t_f = t$ and $x_f = x$, respectively.
$$ \int_{x_0}^{x} dx(t) = \overline{v} \int_{0}^{t} dt, $$
$$ x  x_0 = \overline{v}t,$$
leading to
$$\boxed{x = x_0 + \overline{v}t}$$ (1) 
The above equation allows us to compute the final position $x$ of an object given its initial position $x_0$, constant velocity $\overline{v}$ and the elapsed time $t$.
Graphical Methods

The definition of velocity as the rate of change of position tells us that the slope of the $x$$t$ graph is the velocity.

Also, we deduce that the area under the $v$$t$ graph is the displacement:
$$ \int_{x_0}^{x_f} dx(t) = \int_{t_i}^{t_f} v(t)dt. $$
$$ x_fx_0 = \int_{t_i}^{t_f} v(t)dt.$$
Two runners in a marathon, $A$ and $B$, start running at the same time. The initial position of $A$ is $0~$m and his velocity is $+20~$km/h. The initial position of $B$ is $150~$km and his velocity is $30~$km/h. Determine the time and position at which they meet
(a) using the kinematic equations.
(b) using a graphical method.
Solution (click to expand)
Two balls start out moving to the right with constant velocities of $5.0~$m/s and $4.0~$m/s. The slow ball starts first and the other $4.0~$s later. Determine the position of the balls when they meet
(a) using the kinematic equations.
(b) using a graphical method.
Solution (click to expand)
$t=20~$s.
Uniformly Accelerated Motion
 In a uniformly accelerated motion, the object moves with a constant acceleration. In other words, the $v$$t$ graph is a straight line with a constant slope. In equal intervals of time $\Delta t$, the velocity changes by the same amount $\Delta v$.
Equations of Motion
 We recall the definition of acceleration,
$$ a(t)=\frac{dv(t)}{dt}. $$

We now express $dx(t)$ as the subject and integrate both sides:
$$ \int_{u}^{v} dv(t) = \int_{t_i}^{t_f} a(t)dt. $$
where $u$ and $v$ are the initial and final velocities corresponding to the initial and final times $t_i$ and $t_f$, respectively.

Since we are dealing with uniformlyaccelerated motion, $a(t) = \overline{a}$ is a constant and can be taken out of the integral. Furthermore, we start our clock when motion starts: i.e. $t_i=0$ and let the final time be $t_f = t$.
$$ \int_{u}^{v} dv(t) = \overline{a} \int_{0}^{t} dt, $$
$$ v  u = \overline{a}t,$$
leading to
$$\boxed{v = u + \overline{a}t}$$ (2) 
The above equation allows us to compute the final velocity $v$ of an object given its initial velocity $u$, constant acceleration $\overline{a}$ and the elapsed time $t$.

Further, we now invoke the definition of the velocity,
$$ v(t)=\frac{dx(t)}{dt},$$
substitute equation (2) and integrate both sides to obtain:
$$ \int_{t_0}^t (u + \overline{a}t )dt = \int_{x_0}^x dx(t)$$
$$ \left[ ut + \overline{a}\frac{t^2}{2} \right]_{t_0}^t = x  x_0. $$
Finally, assuming we start our clock when motion starts (i.e. $t_0=0$):
$$\boxed{x = x_0 + ut + \frac{1}{2}\overline{a}t^2}$$ (3) which allows us to compute the final position $x$ of an object given its initial position $x_0$, initial velocity $u$, constant acceleration $\overline{a}$ and the elapsed time $t$ (in a uniformlyaccelerated motion).

Now, let us derive another relationship by eliminating the time $t$ from equations (2) and (3). We first express $t$ as the subject in equation (2):
$$ t = \frac{vu}{\overline{a}} $$
which is then substituted into equation (3) to obtain
$$ x = x_0 + u\left(\frac{vu}{\overline{a}}\right) + \frac{1}{2}\overline{a}\left(\frac{vu}{\overline{a}}\right)^2 $$
Simplifying and rearranging, we obtain:
$$\boxed{v^2 = u^2 + 2 \overline{a} (x  x_0)}$$ (4) 
Equations (2), (3) and (4) will be used to solve uniformlyaccelerated motion problems.
Graphical Methods

The definition of acceleration as the rate of change of velocity tells us that the slope of the $v$$t$ graph is the acceleration.

Also, we deduce that the area under the $a$$t$ graph is the change in velocity $\Delta v$:
$$ \int_{u}^{v} dv(t) = \int_{t_i}^{t_f} a(t)dt. $$
$$ \Delta v = vu = \int_{t_i}^{t_f} a(t)dt.$$
Suppose a car accelerates from rest at a constant acceleration of $26.0~$m/s$^2$ for $5.56~$s. How far does it travel in this time? Solve the question with the help of a $v$$t$ graph.
Solution (click to expand)
x=402 m.
A car in rectilinear motion has an initial velocity of $3.0~$m/s at time $t=0~$s and accelerates uniformly at $2.0~$m/s$^2$. At $t=10~$s,
(a) what is the velocity of the car?
(b) what is the displacement of the car?
Solution (click to expand)
An airplane lands with an initial velocity of $70.0~$m/s and then accelerates opposite to the motion at $1.50~$m/s$^2$ for $40.0~$s. What is its final velocity?
Solution (click to expand)
Ans. $10.0~$m/s.
Two cars $A$ and $B$ are travelling in the same direction. $A$ and $B$ are travelling at constant speeds of $40~$m/s and $25~$m/s, respectively. At time $t=0~$s, $A$ overtakes $B$ which immediately accelerates uniformly for $20~$s to reach a constant speed of $50~$m/s. Calculate
(a) the distance car $A$ travels in the first $20~$s.
(b) the acceleration of car $B$ in the first $20~$s.
(c) the distance car $B$ travels in the first $20~$s.
(d) the additional time (beyond $t=20.0~$s) for car $B$ to catch up with car $A$.
(e) the distance each car has travelled when they meet.
(f) the maximum distance between the cars before car $B$ catches up with car $A$.
Solution (click to expand)
On dry concrete, a car can accelerate opposite to the motion at a rate of $7.00~$m/s$^2$, whereas on wet concrete it can accelerate opposite to the motion at only $5.00~$m/s$^2$. Find the distances necessary to stop a car moving at $30.0~$m/s (about $110~$km/h) on
(a) dry concrete and
(b) wet concrete.
(c) Repeat both calculations and find the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of $0.500~$s to get his foot on the brake.
Solution (click to expand)
(a)x=64.3m on dry concrete.
(b)x=90.0m on wet concrete.
Repeat
(a) to be 64.3 m + 15.0 m = 79.3 m
(b) to be 90.0 m + 15.0 m = 105 m when wet.