Q.If X = 10 - 3t + 6t^2, what is acceleration?

Uniform Acceleration Kinematics: From Intuition to Equations

Imagine you're riding a bicycle down a straight, gentle slope. You don't pedal — you just let gravity pull you. At first you're moving slowly, but soon you're flying. Your speed increases by the same amount every second. That's the core idea: uniform acceleration means your velocity changes by a fixed amount in each equal interval of time.

If you gain 2 m/s every second, your acceleration is $2 \, \text{m/s}^2$. The "per second per second" is the key — it's a rate of change of a rate of change.

The Precise Statement

Uniform (or constant) acceleration means the acceleration vector does not change with time. In one dimension (straight-line motion), this gives us five interconnected quantities:

• $u$ — initial velocity (at $t = 0$)
• $v$ — final velocity (at time $t$)
• $a$ — constant acceleration
• $t$ — time elapsed
• $s$ — displacement (change in position)

These are linked by three equations, often called the equations of motion for constant acceleration:

Each equation omits one variable — the first omits $s$, the second omits $v$, the third omits $t$. This is deliberate: you pick the equation that matches the quantities you know and the one you need.

Where Do These Come From?

Start with the definition of acceleration: $a = \frac{\Delta v}{\Delta t}$. For constant $a$, this becomes $a = \frac{v - u}{t}$. Rearranging gives $v = u + at$ — the first equation.

Now, for constant acceleration, velocity changes linearly with time. The average velocity is simply the arithmetic mean of initial and final velocities: $\frac{u + v}{2}$. Displacement is average velocity times time:

Substitute $v = u + at$ into this:

That's the second equation.

The third equation comes from eliminating $t$. From $v = u + at$, we get $t = \frac{v - u}{a}$. Substitute into $s = \frac{u+v}{2} \cdot t$:

Multiply both sides by $2a$: $v^2 - u^2 = 2as$, or $v^2 = u^2 + 2as$.

A Worked Example

A car starts from rest and accelerates uniformly at $3 \, \text{m/s}^2$ for $5$ seconds. Find its final velocity and the distance travelled.

Given: $u = 0$, $a = 3$, $t = 5$

Final velocity: $v = u + at = 0 + 3 \times 5 = 15 \, \text{m/s}$

Distance: $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 3 \times 25 = 37.5 \, \text{m}$

You could also use $v^2 = u^2 + 2as$ to check: $15^2 = 0 + 2 \times 3 \times s$ gives $225 = 6s$, so $s = 37.5 \, \text{m}$. Consistent.

Common Pitfalls …