The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between t = 0 to t = 4s?

Solution

Physics sir · 3 steps

On a v-t graph the distance is the area under the line, so split it into a triangle and a rectangle.

1. Read the graph. From t = 0 to t = 2 s the velocity rises steadily from 0 to 10 m s⁻¹. From t = 2 s to t = 4 s it stays at 10 m s⁻¹. The velocity never goes negative, so the distance equals the area under the graph.
2. Split the area. The part from O to A is a triangle with base 2 s and height 10 m s⁻¹. The part from A to B is a rectangle with width 2 s and height 10 m s⁻¹.

   s = 1/2 × 2 × 10 + 2 × 10 = 10 + 20 = 30 m

3. So. The object covers 30 m between t = 0 and t = 4 s, which is option (b).

The idea to keep. Area under a v-t graph gives displacement. Break the area into triangles and rectangles, and stop exactly at the time limits the question asks for.

Answer from the NTA final answer key. Solution by PrepWiser.