Calculating Acceleration and Uniform Motion

Calculating Average Acceleration

To calculate average acceleration, we use the formula:

Where $a$ is acceleration ($m/s^2$), $\Delta v$ is the change in velocity ($m/s$), and $t$ is the time taken ($s$). The change in velocity is calculated by subtracting the initial velocity from the final velocity. The time taken is the duration over which this change occurs.

Worked Example 1: Average Acceleration

Scenario: A high-speed train increases its velocity from $30 \, m/s$ to $42 \, m/s$ in $60 \, s$. Calculate the acceleration.

• Step 1: List knowns. Initial velocity ($u$) = $30 \, m/s$, Final velocity ($v$) = $42 \, m/s$, time taken ($t$) = $60 \, s$.

• Step 2: Calculate $\Delta v$. $42 \, m/s - 30 \, m/s = 12 \, m/s$.

• Step 3: State the equation. $a = \frac{\Delta v}{t}$.

• Step 4: Substitute. $a = \frac{12}{60}$.

• Step 5: Solve and state units. $a = 0.2 \, m/s^2$.

Uniform Acceleration

• You might think a car braking to a stop is doing the opposite of accelerating, but in physics, slowing down is just acceleration in the opposite direction.
• When an object's velocity changes by the exact same amount every second, it experiences uniform acceleration.
• This constant rate can be interpreted from a velocity-time graph, where a straight-line gradient indicates constant acceleration, and a horizontal line indicates zero acceleration.
• If an object is slowing down, it experiences deceleration. Because acceleration is a vector, deceleration must be substituted into calculations as a negative number.

The Uniform Acceleration Equation

For uniform acceleration, the AQA equation sheet provides:

Where $v$ is final velocity ($m/s$), $u$ is initial velocity ($m/s$), $a$ is acceleration ($m/s^2$), and $s$ is distance ($m$). Note that the symbol $s$ stands for distance; it does NOT represent speed.

Worked Example 2: Calculating Distance

Scenario: A car accelerates from $5 \, m/s$ to $15 \, m/s$ at a constant rate of $2 \, m/s^2$. Calculate the distance travelled.

• Step 1: Identify knowns. $u = 5 \, m/s$, $v = 15 \, m/s$, $a = 2 \, m/s^2$.

• Step 2: State the equation. $v^2 - u^2 = 2as$.

• Step 3: Rearrange. To find distance ($s$), rearrange to $s = \frac{v^2 - u^2}{2a}$.

• Step 4: Substitute. $s = \frac{15^2 - 5^2}{2 \times 2}$.

• Step 5: Calculate. $s = \frac{225 - 25}{4} = \frac{200}{4} = 50 \, m$.

Worked Example 3: Braking to a Stop

Exam questions often use words instead of numbers. "From rest" means initial velocity is $0 \, m/s$, and "to a stop" means final velocity is $0 \, m/s$. Scenario: A cyclist at $10 \, m/s$ brakes and decelerates at $2.5 \, m/s^2$ until they stop. Calculate the distance.

• Step 1: Identify knowns. $u = 10 \, m/s$, $v = 0 \, m/s$ (because they stop), $a = -2.5 \, m/s^2$ (negative for deceleration).

• Step 2: Substitute into $s = \frac{v^2 - u^2}{2a}$. .

• Step 3: Calculate the squares. $s = \frac{0 - 100}{-5}$.

• Step 4: Solve. $s = \frac{-100}{-5} = 20 \, m$.