question_answer

Two masses A and B are moving in the same straight line. A moves with a uniform velocity\[\text{11 m}{{\text{s}}^{-\text{1}}}\]. B starts from rest at the instant it is 52.5 m ahead of A and moves with a uniform acceleration of\[\text{1m}{{\text{s}}^{-\text{2}}}\]. When will A catch B? Explain the double answer.

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.