A car moves for 2 minutes at a constant velocity of 6 m/s, then accelerates at 1 m/s2 for 6 seconds. Calculate the total displacement of the car.

The car moves for $2 \text{ minutes}$ at constant velocity, and then for $6 \text{ seconds}$ with constant acceleration.
First, convert $2 \text{ minutes}$ to seconds:

$$t_1 = 2 \text{ minutes} \times 60 \text{ s/minute} = 120 \text{ s}$$

The total time interval is $120 \text{ s} + 6 \text{ s} = 126 \text{ s}$.

During this phase, the velocity is constant at $v = 6 \text{ m s}^{-1}$.
On a velocity-time graph, this is represented by a horizontal line from $t=0$ to $t=120 \text{ s}$ at $v=6 \text{ m s}^{-1}$. The area under this segment is a rectangle.

$$s_1 = \text{velocity} \times \text{time}$$ $$s_1 = 6 \text{ m s}^{-1} \times 120 \text{ s} = 720 \text{ m}$$

At the start of this phase (at $t=120 \text{ s}$), the initial velocity is $u = 6 \text{ m s}^{-1}$.
The acceleration is $a = 1 \text{ m s}^{-2}$ for $t_2 = 6 \text{ s}$.
First, calculate the final velocity $v$ at $t=126 \text{ s}$:

$$v = u + at_2$$ $$v = 6 \text{ m s}^{-1} + (1 \text{ m s}^{-2} \times 6 \text{ s})$$ $$v = 6 \text{ m s}^{-1} + 6 \text{ m s}^{-1} = 12 \text{ m s}^{-1}$$

On a velocity-time graph, this phase is represented by a straight line sloping upwards from $(120 \text{ s}, 6 \text{ m s}^{-1})$ to $(126 \text{ s}, 12 \text{ m s}^{-1})$. The area under this segment is a trapezium.
The displacement $s_2$ is the area of this trapezium:

$$s_2 = \frac{1}{2}(u + v)t_2$$ $$s_2 = \frac{1}{2}(6 \text{ m s}^{-1} + 12 \text{ m s}^{-1}) \times 6 \text{ s}$$ $$s_2 = \frac{1}{2}(18 \text{ m s}^{-1}) \times 6 \text{ s}$$ $$s_2 = 9 \text{ m s}^{-1} \times 6 \text{ s} = 54 \text{ m}$$

The total displacement is the sum of the displacements from both phases.

$$S_{\text{total}} = s_1 + s_2$$ $$S_{\text{total}} = 720 \text{ m} + 54 \text{ m} = 774 \text{ m}$$

The total displacement of the car is $\text{774 m}$.