Kinematic Equations

Table of Contents

---

• Introduction
• Uniform Motion
  • Equations of Motion
  • Graphical Methods
• Uniformly Accelerated Motion
  • Equations of Motion
  • Graphical Methods

---

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:

  $$ \int_{x_0}^{x_f} dx(t) = \int_{t_i}^{t_f} v(t)dt. $$

  $$ x_f-x_0 = \int_{t_i}^{t_f} v(t)dt.$$

Solution (click to expand)

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 uniformly-accelerated 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 uniformly-accelerated 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):

  $$ t = \frac{v-u}{\overline{a}} $$

  which is then substituted into equation (3) to obtain

  $$ x = x_0 + u\left(\frac{v-u}{\overline{a}}\right) + \frac{1}{2}\overline{a}\left(\frac{v-u}{\overline{a}}\right)^2 $$

  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 uniformly-accelerated 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 $\Delta v$:

  $$ \int_{u}^{v} dv(t) = \int_{t_i}^{t_f} a(t)dt. $$

  $$ \Delta v = v-u = \int_{t_i}^{t_f} a(t)dt.$$

Solution (click to expand)

x=402 m.

Solution (click to expand)

Solution (click to expand)

Ans. $10.0~$m/s.

Solution (click to expand)

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.