- AvA > vB > vC
- BvA < vB < vC
- CvA = vB = vC
- D$\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$
$\text{v}_\text{A}>\text{v}_\text{B}>\text{v}_\text{C}$
$\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$
Explanation:
Since the speed of light is a universal constant,
$\text{v}_\text{A}=\text{v}_\text{B}=\text{v}_\text{C}=3\times 10^8\text{m/s}$
$\text{v}_\text{B}=\frac{1}{2}(\text{u}_\text{A}+\text{u}_\text{C})$ This expression also implies that vA = vB = vC
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