The velocity-time graph of a particle moving along a straight line is shown in the figure given below. The displacement of the particle in 5 seconds is




(A) 0.5m(B) 1m(C) 2m(D) 4m

To determine the displacement of the particle in 5 seconds based on the velocity-time graph, we need to find the area under the velocity-time curve within that time interval.

Since the velocity-time graph represents the rate of change of displacement with respect to time, finding the area under the curve will give us the total displacement.

In the given figure, we can see that the graph consists of two parts: a rectangle and a triangle.

The rectangle has a base of 5 seconds (time) and a height of 2 m/s (velocity). Therefore, the area of the rectangle is:

Area_rect = base × height = 5 s × 2 m/s = 10 m

The triangle has a base of 5 seconds (time) and a height of 2 m/s (velocity). Therefore, the area of the triangle is:

Area_triangle = 0.5 × base × height = 0.5 × 5 s × 2 m/s = 5 m

To find the total displacement, we add the areas of the rectangle and the triangle:

Total displacement = Area_rect + Area_triangle = 10 m + 5 m = 15 m

Therefore, the displacement of the particle in 5 seconds is 15 meters.

None of the answer options provided (A) 0.5m, (B) 1m, (C) 2m, or (D) 4m match the calculated displacement of 15 meters.