Kinematics

Kinematics: The Mathematics of Motion, the Kenyan Way!

Habari mwanafunzi! Ever been in a matatu on the Thika Superhighway, feeling the push back into your seat as it accelerates, and then the lurch forward as the driver brakes suddenly? Or have you watched our legendary athletes like Eliud Kipchoge and Faith Kipyegon, and marvelled at their incredible, consistent speed? What you're experiencing and observing is Kinematics in action! It's the branch of mechanics that describes the motion of objects without worrying about what causes the motion (we'll leave the forces for Dynamics). Today, we're going to become masters of describing motion using the language of mathematics. Let's begin!

The Core Ingredients: The Language of Motion

Before we can solve any problems, we need to understand the key vocabulary. Think of these as the main characters in our story of motion.

• Displacement (s): This is not just distance! Displacement is the shortest distance from the starting point to the ending point, in a specific direction. It's a vector. Distance, on the other hand, is the total path covered and is a scalar.

  Imagine walking from your classroom to the school's dining hall (DH). You might walk along the corridors, go around the library, and then reach the DH. The total path you walked is the distance. The straight line from the classroom door directly to the DH door is the displacement.

  
  Classroom                                  Library
      o-----------------------------------------+
      |                                         |
      |                                         |  <-- This whole path is DISTANCE
      |                                         |
      +-----------------------------------------o
                                              Dining Hall
  
      o.........................................o
  Classroom                                 Dining Hall
         <-- This straight arrow is DISPLACEMENT
  

• Velocity (v): This is speed in a given direction. Like displacement, it's a vector. The speedometer in a car shows speed (e.g., 80 km/h), but velocity tells us the speed and the direction (e.g., 80 km/h towards Mombasa).
  • Initial Velocity (u): The velocity at the start.
  • Final Velocity (v): The velocity at the end.
• Acceleration (a): This is the rate at which velocity changes. If a boda boda rider opens the throttle, they accelerate. If a bus driver applies the brakes for a speed bump, they decelerate (negative acceleration). It's also a vector.
• Time (t): This one's straightforward! It's the duration over which the motion occurs. It's a scalar.

The Mechanic's Toolkit: The SUVAT Equations

For motion with constant acceleration, we have a set of powerful formulas. We often call them the 'SUVAT' equations, named after the variables they contain: s, u, v, a, t. These are your best friends in kinematics!

• v = u + at       (Doesn't involve displacement, s)
• s = ut + ½at²   (Doesn't involve final velocity, v)
• v² = u² + 2as   (Doesn't involve time, t)
• s = ½(u + v)t   (Doesn't involve acceleration, a)

Let's Get Our Hands Dirty: Worked Examples

Example 1: The Ngong Road Matatu

A matatu, starting from rest at a bus stop, accelerates uniformly at 2 m/s² for 6 seconds along a straight road. Calculate its final velocity and the distance it has covered.

Let's break it down, step-by-step.

--- Step 1: List what we know (our variables) ---
Displacement (s) = ?
Initial Velocity (u) = 0 m/s  (because it starts "from rest")
Final Velocity (v) = ?
Acceleration (a) = 2 m/s²
Time (t) = 6 s

--- Step 2: Find the final velocity (v) ---
We need an equation with v, u, a, and t. The first one is perfect!
v = u + at
v = 0 + (2)(6)
v = 12 m/s

So, the final velocity of the matatu is 12 m/s.

--- Step 3: Find the distance covered (s) ---
We can use a few equations here, but let's use the one that doesn't need 'v'.
s = ut + ½at²
s = (0)(6) + ½(2)(6)²
s = 0 + ½(2)(36)
s = 1 * 36
s = 36 m

The matatu covered a distance of 36 meters. Easy, right?

Example 2: The Mango Drop

A ripe mango drops from a branch that is 5 meters above the ground. Ignoring air resistance, how long does it take to hit the ground? (Take acceleration due to gravity, g = 9.8 m/s²).

Motion under gravity is a classic kinematics problem. The key is that the acceleration is always 'g' acting downwards.

--- Step 1: List the variables (Let's take 'down' as the positive direction) ---
Displacement (s) = +5 m
Initial Velocity (u) = 0 m/s  (because it "drops", it wasn't thrown)
Final Velocity (v) = ? (We don't need it for this question)
Acceleration (a) = +9.8 m/s² (It's positive because it acts in our chosen positive direction)
Time (t) = ?

--- Step 2: Choose the right equation ---
We have s, u, a, and we need t. The perfect formula is:
s = ut + ½at²

--- Step 3: Substitute and solve for t ---
5 = (0)(t) + ½(9.8)(t²)
5 = 0 + 4.9t²
5 = 4.9t²
t² = 5 / 4.9
t² ≈ 1.02
t = √1.02
t ≈ 1.01 s

It takes the mango approximately 1.01 seconds to hit the ground.

Level Up: When Acceleration Isn't Constant (Calculus Kinematics)

In Advanced Mathematics, we often deal with situations where acceleration changes. When this happens, our trusty SUVAT equations don't work. We need a more powerful tool: Calculus! The relationship is beautiful and simple:

• Velocity is the rate of change of displacement: v = ds/dt
• Acceleration is the rate of change of velocity: a = dv/dt = d²s/dt²

This means if you have an equation for displacement (s) in terms of time (t), you can differentiate it to find velocity, and differentiate again to find acceleration!

Calculus Example: The Boda Boda Weave

A boda boda moves along a straight path such that its displacement, s (in meters), from a junction is given by the equation: s = t³ - 6t² + 5, where t is time in seconds. Find its velocity and acceleration when t = 4 seconds.

--- Step 1: We are given the displacement equation ---
s = t³ - 6t² + 5

--- Step 2: Differentiate s with respect to t to find velocity (v) ---
v = ds/dt
v = d/dt (t³ - 6t² + 5)
v = 3t² - 12t

--- Step 3: Differentiate v with respect to t to find acceleration (a) ---
a = dv/dt
a = d/dt (3t² - 12t)
a = 6t - 12

--- Step 4: Now, substitute t = 4 into our new equations for v and a ---
Velocity at t = 4:
v = 3(4)² - 12(4)
v = 3(16) - 48
v = 48 - 48
v = 0 m/s
(This means for a brief instant at 4 seconds, the boda boda momentarily stopped!)

Acceleration at t = 4:
a = 6(4) - 12
a = 24 - 12
a = 12 m/s²
(Even though its velocity was zero, it was accelerating strongly!)

See how calculus gives us a moment-by-moment picture of the motion? That's its power!

Your Turn to Be the Expert!

You've seen the concepts, the formulas, and the examples. From matatus to mangoes, kinematics is everywhere. The key is to read the question carefully, list your variables, pick the right tool (SUVAT or calculus), and solve with confidence.

Challenge Question: A cheetah spots a gazelle 100m away and starts from rest, accelerating at a constant 4 m/s². The gazelle, seeing the cheetah, starts running away at the same instant with a constant velocity of 10 m/s. Will the cheetah catch the gazelle? If so, after how long and how far has the cheetah run?

Give it a try! The world of motion is now yours to describe. Keep practising, stay curious, and you'll master this in no time. Kazi nzuri!