A) 6 s
B) 8 s
C) 10 s
D) The boy cannot catch the bus
Correct Answer: B
Solution :
[b] The bus moves a distance, \[x=\,\frac{1}{2}\,a{{t}^{2}}\] |
The man moves a distance, \[x+48=v\times t\] |
\[\frac{1}{2}\,\times 1\times \,{{t}^{2}}+48=\,10t\] |
\[{{t}^{2}}20t+96=0\] |
\[(t12)\text{ (}t8)=0\] |
\[t=12,\text{ }s,\text{ }t=8s\] |
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