A) 4.0 m/s
B) 5.0 m/s
C) 5.5m/s
D) 4.8m/s
Correct Answer: A
Solution :
[a] Let s be the distance covered, then |
\[{{t}_{1}}=\frac{s/2}{3}=\frac{s}{6}\] |
\[{{t}_{1}}=\frac{s}{2}=4.5{{t}_{2}}+7.5{{t}_{3}}=12{{t}_{2}}\] \[(\because {{t}_{2}}={{t}_{3}})\] |
\[\therefore \operatorname{Average} speed=\frac{Total distance }{Total time}\] |
\[=\frac{s}{\frac{s}{6}+\frac{s}{24}+\frac{s}{24}}\] = 4 m/s |
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