The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is ̱_______ km .

Solution

Physics sir · 4 steps

Distance is the area under the v-t graph: a trapezium for the first 2 s, then a rectangle up to 30.5 s.

1. Read the graph. The velocity rises from 200 m s⁻¹ at t = 0 to 400 m s⁻¹ at t = 2 s (point A). After that it stays at 400 m s⁻¹. The time axis is not drawn to scale, so use the marked values and not the spacing.
2. First 2 seconds. This part is a trapezium with parallel sides 200 and 400 and width 2 s. Its area is 1/2(200 + 400) × 2 = 600 m.
3. Constant speed part. From 2 s to 30.5 s, this part is a rectangle of height 400 m s⁻¹ and width 30.5 - 2 = 28.5 s. Its area is 400 × 28.5 = 11400 m.

   s = 600 m + 11400 m = 12000 m

4. So. The airplane covers 12000 m = 12 km, which is option (a).

The idea to keep. On a v-t graph, split the area into simple shapes wherever the slope changes, add the areas, and convert units only at the end. Always read time values from the labels, not from the spacing on the axis.

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