Introduction to Kinematics and Equations of Motion

In mathematical terms, deceleration is simply treated as negative acceleration. When solving problems, it is vital to assign a direction to your variables; for example, if moving right is positive, then slowing down while moving right requires a negative acceleration value.

Uniform Acceleration and the SUVAT Equations

When an object undergoes uniformly accelerated motion, it means the acceleration remains constant over time. For these specific scenarios, we use a set of five equations, often called the SUVAT equations.

The variables are defined as:

• $s$: Displacement
• $u$: Initial velocity (at $t = 0$)
• $v$: Final velocity
• $a$: Acceleration
• $t$: Time

The equations are:

1. $v = u + at$
2. $v^2 = u^2 + 2as$
3. $s = ut + \frac{1}{2}at^2$
4. $s = \frac{u + v}{2}t$
5. $s = vt - \frac{1}{2}at^2$

Exam Tip: These equations are only applicable when acceleration is constant. If the acceleration changes during the motion, you cannot use these formulas directly for the entire duration.

Graphical Analysis of Motion

Graphs are powerful tools for visualizing and solving kinematics problems. The type of graph you use determines what information you can extract.

Velocity-Time ($v$-$t$) Graphs

The velocity-time graph is perhaps the most versatile tool in kinematics.

• Gradient (Slope): The slope of the line on a $v$-$t$ graph represents the acceleration. A straight line indicates uniform acceleration, while a curved line indicates non-uniform acceleration.
• Area Under the Curve: The total area between the plotted line and the time axis represents the displacement traveled by the object.

Displacement-Time ($s$-$t$) Graphs

• Gradient: The slope of an $s$-$t$ graph represents the velocity.
• Stationary Objects: A horizontal line on an $s$-$t$ graph indicates the object is stationary.

Problem-Solving Strategy

When faced with a kinematics word problem, follow this systematic approach to ensure accuracy:

1. List your variables: Write down the values for $s, u, v, a,$ and $t$.
2. Identify the knowns and unknowns: Determine which values are given and which one you need to find.
   • Note: "Starting from rest" means $u = 0$. "Coming to a stop" means $v = 0$.
3. Account for direction: Ensure deceleration is treated as a negative value.
4. Select the equation: Choose the SUVAT equation that contains your known quantities and your target unknown.
5. Rearrange and solve: Algebraically isolate the unknown variable before plugging in the numbers.

Worked Example

Question: A car applies brakes for $3.00\text{ s}$, and its velocity decreases to $4.50\text{ ms}^{-1}$. The deceleration has a magnitude of $2.70\text{ ms}^{-2}$. What was the original speed of the car?

Solution:

1. Identify knowns:
   • $v = 4.50\text{ ms}^{-1}$
   • $t = 3.00\text{ s}$
   • $a = -2.70\text{ ms}^{-2}$ (negative because it is deceleration)
   • $u = ?$ (this is what we want to find)
2. Select equation: Use $v = u + at$.
3. Rearrange for $u$: $u = v - at$
4. Substitute and solve: $u = 4.50 - (-2.70 \times 3.00)$ $u = 4.50 + 8.10$ $u = 12.6\text{ ms}^{-1}$

By combining algebraic equations with graphical interpretation, you can master the description of motion in any one-dimensional system.