A particle starts from rest at t=0 and moves in the positive x-direction. The given figure shows the velocity-time graph. What is the position of the particle at t=5s?

The position of a particle at a given time is the area under its velocity-time graph from $t=0$ to that time. Since the particle starts from the origin ($x=0$ at $t=0$), its position at $t=5$ s is the total displacement from $t=0$ to $t=5$ s.

The graph can be divided into three distinct regions:

• Region 1: From $t=0$ s to $t=2$ s, a triangle.
• Region 2: From $t=2$ s to $t=4$ s, a rectangle.
• Region 3: From $t=4$ s to $t=5$ s, a rectangle.

• Region 1 (0 s to 2 s): This is a triangle with base 2 s and height 2 m/s. $$\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ s}) \times (2 \text{ m/s}) = 2 \text{ m}$$
• Region 2 (2 s to 4 s): This is a rectangle with width $(4-2) = 2$ s and height 2 m/s. $$\text{Area}_2 = \text{width} \times \text{height} = (2 \text{ s}) \times (2 \text{ m/s}) = 4 \text{ m}$$
• Region 3 (4 s to 5 s): This is a rectangle with width $(5-4) = 1$ s and height 3 m/s. $$\text{Area}_3 = \text{width} \times \text{height} = (1 \text{ s}) \times (3 \text{ m/s}) = 3 \text{ m}$$

The total position $x(5)$ is the sum of the areas of these three regions:

The position of the particle at $t=5$ s is 9 m.

The final answer is $\text{9 m}$.