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.
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