Question 15: Motion in 1D

Question

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 .

Accepted Answer

The distance covered by the airplane in the first 30.5 seconds can be determined by calculating the area under the velocity-time graph.
The formula for distance in this scenario is:
$ 	ext{Distance} = 	ext{Area under the graph} $
Calculating the individual areas:
$ 300 \, 	ext{m/s} 	imes 2 \, 	ext{s} = 600 \, 	ext{m} $
$ 400 \, 	ext{m/s} 	imes 28.5 \, 	ext{s} = 11400 \, 	ext{m} $
Adding these values gives the total distance:
$ 	ext{Total distance} = 600 \, 	ext{m} + 11400 \, 	ext{m} = 12000 \, 	ext{m} $
Thus, the distance covered by the airplane in the first 30.5 seconds is $ 12000 \, 	ext{m} $, which is equivalent to 12 km.

Answer

12

Answer

6

Answer

3

Answer

9

Motion in 1D

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 .

Choose the correct answer: