A1-3. Uniform Acceleration Motion

Defining Symbols (SUVAT)

To solve kinematics problems involving uniform acceleration, we assign specific letters to represent each variable. (Note: While the IB standard uses SUVAT, many textbooks use alternative symbols like $S$ or $\Delta x$ for displacement, and $v_0$ for initial velocity. They mean the exact same thing.)

Variable Symbol Alternative Symbol Unit
Displacement $s$ $S$ or $\Delta x$ metres ($m$)
Initial Velocity $u$ $v_0$ metres per second ($\text{m/s}$)
Final Velocity $v$ $v$ metres per second ($\text{m/s}$)
Acceleration $a$ $a$ metres per second squared ($\text{m/s}^2$)
Time $t$ $t$ seconds ($s$)

⚠ Crucial Condition

The equations derived below can only be used for motion with UNIFORM (CONSTANT) ACCELERATION.

Deriving the Equations of Motion

In uniform acceleration, velocity changes at a constant rate. We can derive the core equations algebraically and by looking at the area bounded under a velocity-time ($v-t$) graph.

Time (t) Velocity (v) u v 0 t Area = ut Area = ½(v-u)t

Equation 1: Velocity-Time
Starting with the basic equation for acceleration ($a = \dfrac{v-u}{t}$), multiply both sides by $t$ to get $at = v - u$. Rearranging gives:

$$v = u + at$$

Equation 2: Average Velocity & Displacement
The displacement ($s$ or $\Delta x$) is equal to the total trapezoidal area enclosed under the velocity-time graph. Using the area formula for a trapezoid $$ \left(\dfrac{\text{top base} + \text{bottom base}}{2} \times \text{height}\right) $$ we have

$$s = \dfrac{1}{2}(u + v)t$$

Notice that $\dfrac{1}{2}(u+v)$ is simply the average velocity.

Equation 3: Eliminating Final Velocity ($v$)
We can break the trapezoid into a rectangle and a triangle (as seen in the diagram).
Rectangle Area = $ut$
Triangle Area = $\dfrac{1}{2}(v-u)t$. Since $v-u = at$ (from Eq 1), the triangle area is $$ \dfrac{1}{2}(at)t = \dfrac{1}{2}at^2 $$ Adding them together, or substituting Equation 1 into Equation 2, yields:

$$s = ut + \dfrac{1}{2}at^2$$

Equation 4: Eliminating Time ($t$)
Rearrange Equation 1 to solve for time: $t = \dfrac{v-u}{a}$. Substitute this into Equation 2:
$$s = \dfrac{1}{2}(u+v) \left(\dfrac{v-u}{a}\right)$$
$$2as = (v+u)(v-u) = v^2 - u^2$$
Rearranging gives:

$$v^2 = u^2 + 2as$$

Example 1

Problem: An object moving with an initial velocity of 42 m/s accelerates uniformly along a straight path. After 9 s, it reaches a final velocity of 87 m/s. Determine:

  1. The acceleration of the object
  2. The total displacement covered by the object when time reaches $t = 100\text{ s}$

Solution:

  • a) Acceleration ($a$):
    Given: $u = 42\text{ m/s}$, $v = 87\text{ m/s}$, $t = 9\text{ s}$
    $$v = u + at \implies 87 = 42 + a(9) \implies 45 = 9a \implies a = 5\text{ m/s}^2$$ Acceleration = 5 m/s²
  • b) Total Displacement ($s$) at 100 s:
    Given: $u = 42\text{ m/s}$, $a = 5\text{ m/s}^2$, $t = 100\text{ s}$
    $$ \begin{aligned} s &= ut + \dfrac{1}{2}at^2 \implies s = (42)(100) + \dfrac{1}{2}(5)(100)^2 = 4200 + 25000 \\ &= 29200\text{ m} \end{aligned} $$ Total Displacement = 29200 m

Summary of the Four Equations

Write down the three variables you know from the question and the one unknown you are trying to find. Choose the equation that contains exactly those four variables.

  • $$v = u + at$$
  • $$s = \dfrac{1}{2}(u + v)t$$
  • $$s = ut + \dfrac{1}{2}at^2$$
  • $$v^2 = u^2 + 2as$$