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?

Step 1: Understand the relationship between velocity-time graph and position. 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.

Step 2: Divide the area under the graph into geometric shapes from $t=0$ s 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.

Step 3: Calculate the area of each region.

• Region 1 (0 s to 2 s): This is a triangle with base $2$ s and height $2$ m/s. $Area_1 = \frac{1}{2} \times base \times height = \frac{1}{2} \times (2 s) \times (2 m/s) = 2 m$
• Region 2 (2 s to 4 s): This is a rectangle with width $(4-2) = 2$ s and height $2$ m/s. $Area_2 = width \times height = (2 s) \times (2 m/s) = 4 m$
• Region 3 (4 s to 5 s): This is a rectangle with width $(5-4) = 1$ s and height $3$ m/s. $Area_3 = width \times height = (1 s) \times (3 m/s) = 3 m$

Step 4: Sum the areas to find the total position at $t=5$ s. The total position $x(5)$ is the sum of the areas of these three regions: $x(5) = Area_1 + Area_2 + Area_3$ $x(5) = 2 m + 4 m + 3 m = 9 m$

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

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