## Kinematics

When teaching students about kinematics at JC level, I noticed that students tend to find the following challenging:

Hence, I've prepared the following tips to assist them.

###### 1. Understanding the Kinematics Equations

The commonly used symbols for physical quantities in rectilinear motion are:

$a$: acceleration
$v$: final velocity
$u$: initial velocity
$t$: time
$s$: displacement

The four equations below are true for motion in a straight line at constant acceleration.

Equation 1: $a=\frac{v-u}{t}$ or $v=u+at$

The first is derived from finding the gradient of the straight line graph of velocity over time. The second variant is from adding the change in velocity to initial velocity.

Equation 2: $s=\frac{(u+v)}{2}t$

This is derived from finding the area under the velocity-time graph, which takes the form of a trapezium for constant acceleration.

Equation 3: $v^2=u^2+2as$

This equation can be derived from the Work-Energy Theorem, which states that the net work done is equal to the change in kinetic energy.

Equation 4: $s=ut+\frac{1}{2}at^2$

This is obtained from substituting the first equation into the second.

###### 2. Knowing when to use them

The correct equation to use depends on the amount of information given in the question. Using the table below will help identify the equation suitable for the question. Often a question will give three unknown variables and ask you to find a fourth. One of these four is usually {tex}u{/tex} so it leaves us to decide which of the other four physical quantities are not available.

 $a$ $v$ $t$ $s$ $u$ Equation to use × ? ? ? ? $s=\frac{(u+v)}{2}t$ ? × ? ? ? $s=ut+\frac{1}{2}at^2$ ? ? × ? ? $v^2=u^2+2as$ ? ? ? × ? $a=\frac{v-u}{t}$

×: not available      ?: given or need to find

###### 3. Two-Dimensional Motion

When dealing with two-dimensional motion, we have to resolve each of the vectors into two independent components that are perpendicular to each other.

For instance, in the case of a projectile motion starting with a velocity $u$ at an angle $\theta$ to the horizontal, with no horizontal acceleration during flight, the following expressions apply:

 Horizontal component Vertical component Resultant magnitude $a$ 0 $g$ $g$ $u$ $u_x=u\cos\theta$ $u_y=u\sin\theta$ $u$ $v$ $v_x=u\cos\theta$ (no change!) $v_y=u\sin\theta+gt$ $\sqrt{v_x^2+v_y^2}$ $s$ $s_x=u_xt$ $s_y=u_yt+\frac{1}{2}gt^2$ $\sqrt{s_x^2+s_y^2}$