kindly assist.
Asked by Kanhaiya | 16th Sep, 2015, 08:23: PM
Expert Answer:
1. From the graph it is clear that OB' > O'A'.
Thus, A lives closer to the college.
2. The position- time graph of A starts from the origin (t = 6 : 30 ) and B starts from 6 : 34.
This indicates that A started earlier than B. Here, on X-axis 1 unit = 1 min.
Hence, the A started earlier than B by a time interval of OC = 4 min.
3. The speed is represented by the steepnes (slope) of the graph. The position-graph of B is steeper than the position-time graph of A.
Thus, we can conclude that B is faster than A.
4. The time interval of B is less than that of A. That is, OE > OD. This indicates that B reaches earlier at 6 : 42.
5. They cross each other at point K.
1. From the graph it is clear that OB' > O'A'.
Thus, A lives closer to the college.
2. The position- time graph of A starts from the origin (t = 6 : 30 ) and B starts from 6 : 34.
This indicates that A started earlier than B. Here, on X-axis 1 unit = 1 min.
Hence, the A started earlier than B by a time interval of OC = 4 min.
3. The speed is represented by the steepnes (slope) of the graph. The position-graph of B is steeper than the position-time graph of A.
Thus, we can conclude that B is faster than A.
4. The time interval of B is less than that of A. That is, OE > OD. This indicates that B reaches earlier at 6 : 42.
5. They cross each other at point K.
Answered by Yashvanti Jain | 18th Sep, 2015, 12:18: PM
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