Calculate the acceleration for each car.

The formula for constant acceleration starting from rest is $a = \frac{v}{t}$.

For Car A:
Initial velocity $u_A = 0 \text{ m/s}$
Final velocity $v_A = 5 \text{ m/s}$
Time $t_A = 5 \text{ s}$

$$a_A = \frac{v_A}{t_A} = \frac{5 \text{ m/s}}{5 \text{ s}} = 1 \text{ m/s}^2$$

For Car B:
Initial velocity $u_B = 0 \text{ m/s}$
Final velocity $v_B = 3 \text{ m/s}$
Time $t_B = 10 \text{ s}$

$$a_B = \frac{v_B}{t_B} = \frac{3 \text{ m/s}}{10 \text{ s}} = 0.3 \text{ m/s}^2$$

We will calculate velocities at $t = 0, 2.5, 5, 7.5, 10 \text{ s}$ to cover the relevant time range up to $10 \text{ s}$.
The velocity for constant acceleration is $v = u + at$. Since $u=0$, $v=at$.

For Car A ($a_A = 1 \text{ m/s}^2$):

• At $t = 0 \text{ s}$, $v_A = 1 \times 0 = 0 \text{ m/s}$
• At $t = 2.5 \text{ s}$, $v_A = 1 \times 2.5 = 2.5 \text{ m/s}$
• At $t = 5 \text{ s}$, $v_A = 1 \times 5 = 5 \text{ m/s}$
• For $t > 5 \text{ s}$, Car A maintains its velocity of $5 \text{ m/s}$.
• At $t = 7.5 \text{ s}$, $v_A = 5 \text{ m/s}$
• At $t = 10 \text{ s}$, $v_A = 5 \text{ m/s}$

For Car B ($a_B = 0.3 \text{ m/s}^2$):

• At $t = 0 \text{ s}$, $v_B = 0.3 \times 0 = 0 \text{ m/s}$
• At $t = 2.5 \text{ s}$, $v_B = 0.3 \times 2.5 = 0.75 \text{ m/s}$
• At $t = 5 \text{ s}$, $v_B = 0.3 \times 5 = 1.5 \text{ m/s}$
• At $t = 7.5 \text{ s}$, $v_B = 0.3 \times 7.5 = 2.25 \text{ m/s}$
• At $t = 10 \text{ s}$, $v_B = 0.3 \times 10 = 3 \text{ m/s}$

To plot the graphs on the same axes (time on x-axis, velocity on y-axis):

• Car A: The graph starts at $(0,0)$. It is a straight line with a positive slope (representing constant acceleration) from $(0,0)$ to $(5 \text{ s}, 5 \text{ m/s})$. After $5 \text{ s}$, the velocity remains constant at $5 \text{ m/s}$, so the graph becomes a horizontal line from $(5 \text{ s}, 5 \text{ m/s})$ to $(10 \text{ s}, 5 \text{ m/s})$.
• Car B: The graph starts at $(0,0)$. It is a straight line with a positive slope (representing constant acceleration) from $(0,0)$ to $(10 \text{ s}, 3 \text{ m/s})$.

Displacement is the area under the velocity-time graph for the specified time intervals.

For Car A (interval $0 \text{ s}$ to $5 \text{ s}$):
The graph forms a triangle with base $5 \text{ s}$ and height $5 \text{ m/s}$.

$$S_A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \text{ s} \times 5 \text{ m/s} = 12.5 \text{ m}$$

The displacement for Car A is $\boxed{\text{12.5 m}}$.

For Car B (interval $0 \text{ s}$ to $10 \text{ s}$):
The graph forms a triangle with base $10 \text{ s}$ and height $3 \text{ m/s}$.