A) 5 m/s
B) 15 m/s
C) 30 m/s
D) 7.5 m/s
Correct Answer: A
Solution :
Key Idea: \[\text{Average}\,\text{velocity}\,\text{=}\,\frac{\text{Displacement}}{\text{Time}\,\text{taken}}\] Let body move along AB, then BC. Distance travelled in 2s. \[{{\text{S}}_{1}}=15\times 2=30\,m\] Distance travelled in 8s \[{{S}_{2}}=5\times 8=40\,m\] \[\therefore \] Displacement \[AC=\sqrt{{{(30)}^{2}}+{{(40)}^{2}}}\] \[=\sqrt{900+1600}=\sqrt{2500}\] \[=50\,m\] Hence average velocity \[=\frac{\text{displacement}}{\text{time}\,\text{taken}}\] \[=\frac{50}{8+2}=\,5\,m/s\]You need to login to perform this action.
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