A) 12 m/s
B) 14 m/s
C) 15 m/s
D) 16 m/s
Correct Answer: C
Solution :
Let man will catch the bus after 't' sec . So he will cover distance ut. Similarly distance travelled by the bus will be \[\frac{1}{2}a{{t}^{2}}\]. For the given condition \[u\ t=45+\frac{1}{2}a\ {{t}^{2}}\]\[=45+1.25\ {{t}^{2}}\] \[[As\ a=2.5m/{{s}^{2}}]\] Þ \[u=\frac{45}{t}+1.25\ t\] To find the minimum value of u \[\frac{du}{dt}=0\] so we get \[t=6\sec \]then, \[u=\frac{45}{6}+1.25\times 6=7.5+7.5=15m/s\]
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