Answer:
Let A will catch B after time t. Here; \[{{\text{S}}_{\text{1}}}=\text{11t}\]
and \[{{\text{S}}_{\text{2}}}=\frac{1}{2}\times 1\times
{{t}^{2}}=\frac{1}{2}t\]
As, \[{{\text{S}}_{\text{1}}}={{\text{S}}_{\text{2}}}+\text{52}.\text{5}\],
so \[\text{11t}=\frac{1}{2}{{\text{t}}^{\text{2}}}+\text{ 52}.\text{5}\] or \[{{\text{t}}^{\text{2}}}\text{22t
}+\text{1}0\text{5}=0\]
On solving, we shall get t = 7 or 15 s
the has two values because B can also move in opposite direction of A.
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