Question

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

• A. 9
• B. 6
• C. 3
• D. 12

Hint:

To find the distance covered by an object when given a velocity-time graph, calculate the area under the curve. If the graph is a combination of shapes like rectangles and triangles, calculate each area separately and then add them up.

Answer 1

The Correct Option is D

To determine the distance traveled by the airplane within the initial 30.5 seconds, an examination of the velocity-time graph is required. The distance covered is represented by the area beneath this graph.

1. The graph indicates that the airplane's velocity increases uniformly from 0 m/s to 400 m/s over the first 2 seconds. This segment forms a triangular area with the time axis.
2. Subsequently, from 2 seconds to 30.5 seconds, the velocity is sustained at a constant 400 m/s.

The distances for these two intervals are calculated as follows:

1. Distance calculation for the initial 0 to 2 second interval (triangle):
   • Base = 2 seconds
   • Height = 400 m/s
   • Area (Distance) = \(\frac{1}{2} \times 2 \times 400 = 400 \text{ m}\)
2. Distance calculation for the 2 to 30.5 second interval (rectangle):
   • Base = 30.5 s - 2 s = 28.5 seconds
   • Height = 400 m/s
   • Area (Distance) = \(\text{28.5} \times 400 = 11400 \text{ m}\)
3. Total distance calculation:
   • Total Distance = 400 m + 11400 m = 11800 m
   • Conversion to kilometers: \(\frac{11800}{1000} = 11.8 \text{ km}\)

Based on potential rounding or approximation, the closest value is 12 km.

Thus, the final answer is 12 km.