question_answer

A bus is beginning to move with an acceleration of \[\text{1m}/{{\text{s}}^{\text{2}}}\]. A boy, who is 48 m behind the bus, starts running at \[\text{10}\,\text{m}/\text{s}\]. After what time will the boy be able to catch the bus?

Answer:

  Let A and B be the initial positions of the boy and bus respectively. distance travelled by bus, \[x=\frac{1}{2}a{{t}^{2}}\] distance travelled by boy,\[x+48=v\times t\] or \[\frac{1}{2}\times 1\times {{t}^{2}}+48=10t\] or \[{{t}^{2}}-20t+96=0\]or \[(t-12)(t-8)=0\] or \[t=12s\]or \[t=8s\] The boy will cross the bus after 8s. But, again, after 12s the bus will cross him due to accelerated motion.