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Joseph jogs from one end A to the other end B of a straight 30... - Science
Question
Joseph jogs from one end A to the other end B of a straight 300 \text{ m} road in 2 \text{ minutes } 30 \text{ seconds} and then turns around and jogs 100 \text{ m} back to point C in another 1 \text{ minute}. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer
Given:
Distance AB = 300 \text{ m}, Time t_1 = 2 \text{ min } 30 \text{ s} = 150 \text{ s}
Distance BC = 100 \text{ m}, Time t_2 = 1 \text{ min } = 60 \text{ s}
(a) Motion from A to B:
Average Speed:
= \frac{\text{Distance}}{\text{Time}} = \frac{300}{150} = \mathbf{2 \text{ m/s}}Average Velocity:
= \frac{\text{Displacement}}{\text{Time}} = \frac{300}{150} = \mathbf{2 \text{ m/s}}
(b) Motion from A to C:
Total Distance
= 300 + 100 = 400 \text{ m}Total Time
= 150 + 60 = 210 \text{ s}Net Displacement
= 300 - 100 = 200 \text{ m}Average Speed:
= \frac{\text{Total Distance}}{\text{Total Time}} = \frac{400}{210} = \mathbf{1.90 \text{ m/s}}Average Velocity:
= \frac{\text{Net Displacement}}{\text{Total Time}} = \frac{200}{210} = \mathbf{0.95 \text{ m/s}}