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 find the distance covered by the airplane in the first 30.5 seconds, we need to analyze the velocity-time graph. The area under the velocity-time graph gives the distance traveled. 

1. From the graph, we can see that the velocity increases linearly from 0 to 400 m/s in the first 2 seconds, creating a triangle with the time axis.
2. Then, the velocity remains constant at 400 m/s from 2 seconds to 30.5 seconds.

Let's calculate the distance for both sections:

1. Calculate the area of the triangle (0 to 2 seconds):
   • The base of the triangle is 2 seconds.
   • The height of the triangle is 400 m/s.
   • Area of the triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 400 = 400 \text{ m}\)
2. Calculate the area of the rectangle (2 to 30.5 seconds):
   • The base of the rectangle is (30.5 - 2) = 28.5 seconds.
   • The height of the rectangle is 400 m/s.
   • Area of the rectangle = \(\text{base} \times \text{height} = 28.5 \times 400 = 11400 \text{ m}\)
3. Total distance covered:
   • Sum of areas = 400 m + 11400 m = 11800 m
   • Convert to kilometers: \(\frac{11800}{1000} = 11.8 \text{ km}\)

Considering a rounding or approximation in the context, the closest option is 12 km.

Therefore, the correct answer is 12 km.

Answer 2

To calculate the distance covered by the airplane in the first 30.5 seconds, we analyze its motion in two distinct phases:

1. Acceleration Phase (0-2 seconds):
The airplane accelerates uniformly from 200 m/s to 400 m/s.
- Initial velocity (v_i) = 200 m/s
- Final velocity (v_f) = 400 m/s
- Time duration (t) = 2 s

Distance covered:
\[ d_1 = \left(\frac{v_i + v_f}{2}\right) \times t = \left(\frac{200 + 400}{2}\right) \times 2 = 600 \text{ meters} \]

2. Constant Velocity Phase (2-30.5 seconds):
The airplane maintains a constant velocity of 400 m/s.
- Velocity (v) = 400 m/s
- Time duration (t) = 30.5 - 2 = 28.5 s

Distance covered:
\[ d_2 = v \times t = 400 \times 28.5 = 11,400 \text{ meters} \]

3. Total Distance Calculation:
\[ \text{Total distance} = d_1 + d_2 = 600 + 11,400 = 12,000 \text{ meters} \]

4. Unit Conversion:
\[ 12,000 \text{ meters} = 12 \text{ km} \]

Final Answer:
The airplane covers \(\boxed{12}\) kilometers in the first 30.5 seconds.