A) \[u=2.25v\]
B) \[u=3v\]
C) \[u<u<2v\]
D) \[3v<u<4v\]
Correct Answer: C
Solution :
Let side of a square be x To cover x, distance in a side AB, times will be \[=\frac{x}{v}\] Similarly, To cover x distance in a side BC, times will be \[=\frac{x}{2v}\] To cover x. distance in, a side CD, time will be \[=\frac{x}{3v}\] And to cover x, distance in a side DA, time will be \[=\frac{x}{4v}\] \[\because \]\[\text{Average speed}\,\text{(u)=}\frac{\text{Total distance}}{\text{ }\!\!~\!\!\text{ Total time}}\] \[=\frac{(x+x+x+x)}{\frac{x}{v}+\frac{x}{2v}+\frac{x}{3v}+\frac{x}{4v}}\] \[=\frac{48v}{25}\] \[=1.92\,v\] Which lies in the interval \[v<u<2v.\]You need to login to perform this action.
You will be redirected in
3 sec