The problem describes a car's motion in three distinct stages. We will analyze each stage using the principles of kinematics.

Stage 1: Constant velocity motion ($t=0$ to $t=2$ s)

• Initial velocity: $v_0 = 10$ m/s
• Duration: $t_1 = 2$ s
• Acceleration: $a_1 = 0$

Stage 2: Constant acceleration motion ($t=2$ s to $t=5$ s)

• Initial velocity for this stage: $v_1 = v_0 = 10$ m/s
• Duration: $\Delta t_2 = 3$ s
• Acceleration: $a_2 = 1 \text{ m/s}^2$

Stage 3: Constant velocity motion ($t=5$ s to $t=9$ s)

• Duration: $\Delta t_3 = 4$ s
• The velocity is constant at the value reached at the end of Stage 2.

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[Q1] Find the speed of the car during the final phase of constant velocity motion. The speed during the final phase, let's call it $v_2$, is the velocity at the end of the acceleration period (Stage 2). We use the kinematic equation for velocity under constant acceleration.

Derivation: The final velocity $v_f$ is given by $v_f = v_i + at$. Applying this to Stage 2:

$$v_2 = v_1 + a_2 \Delta t_2$$

Substitution:

$$v_2 = 10 \text{ m/s} + (1 \text{ m/s}^2)(3 \text{ s}) = 13 \text{ m/s}$$

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[Q2] Draw the velocity-time (v-t) graph for the entire process. The v-t graph is constructed segment by segment, corresponding to each stage of motion.

1. Stage 1 (0 to 2 s): Constant velocity $v = 10$ m/s. This is a horizontal line.
2. Stage 2 (2 to 5 s): Constant acceleration. The velocity increases linearly from $10$ m/s to $13$ m/s. This is a line with a positive slope.
3. Stage 3 (5 to 9 s): Constant velocity $v = 13$ m/s. This is another horizontal line at a higher velocity.

The key coordinate points $(t, v)$ for the graph are $(0, 10)$, $(2, 10)$, $(5, 13)$, and $(9, 13)$.