The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below:
(a) (A/ B) lives closer to the school than (B/ A)
(b) (A/ B) starts from the school earlier than (B/A)
(c) (A/ B) walks faster than (B/ A)
(d) A and B reach home at the (same/different) time
(e) (A/ B) overtakes (B/ A) on the road (once/twice).
Answers
in first part
_________
1) As the OP <OQ it means A lives closer
2) While T has some infinite value for B but for A, x ==0 and t = 0 So A starts from the school earlier than B
3) Slope of X-t graph of B is greater than A so B walks faster than A
4) The graph shows that both A and B are at home at same time
5) The graph shows that B overtakes only one time to A.
Answer: The answer to this question is-
The two kids are A and B, and their separate residences are P and Q. Their school is O.
a) A lives closer to school than B.
b) A starts from school earlier than B.
c) B walks faster than B.
d) A and B reach home at the same time.
e) B overtakes A once on the road.
Explanation:
a) The school's distance from A's home is shorter than that from B's home since distance OP is shorter than distance OQ. As a result, child A commutes less time to school than does child B.
b) Kid B will be at a finite value on the t-axis if child A is at the origin. As a result, child A arrives at school before child B.
c) It can be seen that child B's slope is higher than child A's slope since the slope of the distance-time graph indicates speed. Child B thus walks more quickly than A.
d) The timeframes for both children A and B to arrive home are the same, as seen in the graph above. Child A and B arrive at home at the same time as a result.
e) Child B moves later than Child A due to Child B's faster than Child A's pace. As a result, kid B only passes child A once on the route.
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