Please answer this question as fast as possible........with full explanation...
Attachments:
Answers
Answered by
0
Hey there!
The graph given there is velocity time graph.
And the total area under the graph will give you the distance (but not displacement,with which we are concerned). We know that displacement is a vector quantity and so it depends on direction too.
So take out the positive area(area above x-axis) and subtract the negative area(area below x-axis) from it.
Now coming to procedure,
OAB is triangle so use the 1/2 ×B×H formula.
BCD too is a triangle but it is below x-axis so you have to subtract it from the positive area.
DEFG is square and will give the positive area.
Net equation : OAB+DEFG -BDC.
which will be 3+1-1= 3 sq. units
The graph given there is velocity time graph.
And the total area under the graph will give you the distance (but not displacement,with which we are concerned). We know that displacement is a vector quantity and so it depends on direction too.
So take out the positive area(area above x-axis) and subtract the negative area(area below x-axis) from it.
Now coming to procedure,
OAB is triangle so use the 1/2 ×B×H formula.
BCD too is a triangle but it is below x-axis so you have to subtract it from the positive area.
DEFG is square and will give the positive area.
Net equation : OAB+DEFG -BDC.
which will be 3+1-1= 3 sq. units
Similar questions