A) 3.6m
B) 28.8m
C) 36.0m
D) 40.0m
Correct Answer: C
Solution :
The area covered between the velocity-time graph and time, axis gives the displacement. Area of \[\Delta ABC=\frac{1}{2}\times AC\times BC\] \[=\frac{1}{2}\times 2\times 3.6=3.6\] Area of \[\square BCDE=BD\times BC\] \[=8\times 3.6=28.8\] Area of \[\Delta DEF=\frac{1}{2}EF\times DE\] \[=\frac{1}{2}\times 2\times 3.6\] \[=3.6\] \[\therefore \] Total area \[=3.6+28.8+3.6=36m\]You need to login to perform this action.
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