Question #7b1de
1 Answer
The displacement - time graph has been shown above.
We know
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This relation shows that displacement time graph is linear when velocity remains same.
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At initial position
#O#
#t=0 and d=0->"coordinate"(0,0)# -
In first phase walking rate is
#100m"/"min# . So in 10 min displacement is 1000m
Hence at# t= 10min# the coordinate of position is#(10,1000)# which is point#A# in the graph. So line OA will represent the displacement - time graph during first 10 min. -
Next 10 min is rest phase so displacement will be same 1000m during this time. So the position after 20 sec will be represnted by the coordinate
#(20,1000)# at B. Hence the line segment joining the points A and B will represent the displacement - time graph during this 10 min period. -
In 3rd phase the rate of walking for return journey of
#500m# was#200m"/"min# . So time required will be#500/200=2.5min# . Thus the coordinate of the position at#t=22.5 min# will be#(22.5,500)# , represented by point#C# . Hence line segment BC represents the displacement-time graph in this phase. -
Finally in last phase the rate of walk is
#150m"/"min# . The displacement is 500m in this phase, So time required will be#500/150~~3.3min# . So total time taken to return in initial position will be#22.5+3.3=25.8min# and the coordinate of this position becomes#(25.8,0)# , the point D. So the line segment CD is the displacement time graph for the last phase,
