The distance -time graph below shows the journey of a car. Convert the the distance -time graph to velocity-time graph and acceleration-time graph.

Velocity is the gradient of the distance-time graph.

• Segment 1 (0 s to 3 s): Points: $(0 \text{ s}, 0 \text{ m})$ and $(3 \text{ s}, 6 \text{ m})$ $$v_1 = \frac{6 \text{ m} - 0 \text{ m}}{3 \text{ s} - 0 \text{ s}} = \frac{6 \text{ m}}{3 \text{ s}} = 2 \text{ m/s}$$
• Segment 2 (3 s to 5 s): Points: $(3 \text{ s}, 6 \text{ m})$ and $(5 \text{ s}, 6 \text{ m})$ $$v_2 = \frac{6 \text{ m} - 6 \text{ m}}{5 \text{ s} - 3 \text{ s}} = \frac{0 \text{ m}}{2 \text{ s}} = 0 \text{ m/s}$$
• Segment 3 (5 s to 9 s): Points: $(5 \text{ s}, 6 \text{ m})$ and $(9 \text{ s}, 24 \text{ m})$ $$v_3 = \frac{24 \text{ m} - 6 \text{ m}}{9 \text{ s} - 5 \text{ s}} = \frac{18 \text{ m}}{4 \text{ s}} = 4.5 \text{ m/s}$$
• Segment 4 (9 s to 12 s): Points: $(9 \text{ s}, 24 \text{ m})$ and $(12 \text{ s}, 24 \text{ m})$ $$v_4 = \frac{24 \text{ m} - 24 \text{ m}}{12 \text{ s} - 9 \text{ s}} = \frac{0 \text{ m}}{3 \text{ s}} = 0 \text{ m/s}$$

The velocity-time graph (with velocity on the y-axis in m/s and time on the x-axis in s) would show:

• A horizontal line at $v = 2 \text{ m/s}$ from $t=0 \text{ s}$ to $t=3 \text{ s}$.
• A horizontal line at $v = 0 \text{ m/s}$ (on the time axis) from $t=3 \text{ s}$ to $t=5 \text{ s}$.
• A horizontal line at $v = 4.5 \text{ m/s}$ from $t=5 \text{ s}$ to $t=9 \text{ s}$.
• A horizontal line at $v = 0 \text{ m/s}$ (on the time axis) from $t=9 \text{ s}$ to $t=12 \text{ s}$.

Acceleration is the gradient of the velocity-time graph. Since the velocity is constant in each segment, the acceleration within each segment is zero.

• Segment 1 (0 s to 3 s): Velocity is constant at $2 \text{ m/s}$. $$a_1 = \frac{\Delta v}{\Delta t} = 0 \text{ m/s}^2$$
• Segment 2 (3 s to 5 s): Velocity is constant at $0 \text{ m/s}$. $$a_2 = \frac{\Delta v}{\Delta t} = 0 \text{ m/s}^2$$
• Segment 3 (5 s to 9 s): Velocity is constant at $4.5 \text{ m/s}$. $$a_3 = \frac{\Delta v}{\Delta t} = 0 \text{ m/s}^2$$
• Segment 4 (9 s to 12 s): Velocity is constant at $0 \text{ m/s}$. $$a_4 = \frac{\Delta v}{\Delta t} = 0 \text{ m/s}^2$$

The acceleration-time graph (with acceleration on the y-axis in m/s$^2$ and time on the x-axis in s) would show:

• A horizontal line at $a = 0 \text{ m/s}^2$ (on the time axis) from $t=0 \text{ s}$ to $t=12 \text{ s}$.

Summary of Graph Conversions:

$$\boxed{ \begin{array}{l} \textbf{Velocity-Time Graph:} \\ \text{• From } t=0 \text{ s to } t=3 \text{ s, velocity is } 2 \text{ m/s.} \\ \text{• From } t=3 \text{ s to } t=5 \text{ s, velocity is } 0 \text{ m/s.} \\ \text{• From } t=5 \text{ s to } t=9 \text{ s, velocity is } 4.5 \text{ m/s.} \\ \text{• From } t=9 \text{ s to } t=12 \text{ s, velocity is } 0 \text{ m/s.} \\ \text{The graph consists of horizontal segments at these velocity values.} \\ \\ \textbf{Acceleration-Time Graph:} \\ \text{• From } t=0 \text{ s to } t=12 \text{ s, acceleration is } 0 \text{ m/s}^2. \\ \text{The graph is a horizontal line along the time axis.} \end{array} }$$