A and B take part in a 100, m race. A runs at | vedclass

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(B) Speed of $A = 5\, km/h = 5 	imes \frac{5}{18} = \frac{25}{18}\, m/s$.Time taken by $A$ to cover $100\, m = \frac{100}{25/18} = 100 	imes \frac{1

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$4.14$

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(B) Speed of $A = 5\, km/h = 5 	imes \frac{5}{18} = \frac{25}{18}\, m/s$.Time taken by $A$ to cover $100\, m = \frac{100}{25/18} = 100 	imes \frac{18}{25} = 72\, s$.Since $A$ gives $B$ a start of $8\, m$,$B$ only needs to cover $100 - 8 = 92\, m$.$A$ beats $B$ by $8\, s$,so $B$ takes $72 + 8 = 80\, s$ to cover the $92\, m$ distance.Speed of $B = \frac{	ext{Distance}}{	ext{Time}} = \frac{92}{80}\, m/s$.To convert into $km/h$,multiply by $\frac{18}{5}$:Speed of $B = \frac{92}{80} 	imes \frac{18}{5} = 1.15 	imes 3.6 = 4.14\, km/h$.

$A$ and $B$ take part in a $100\, m$ race. $A$ runs at $5\, km/h$. $A$ gives $B$ a start of $8\, m$ and still beats him by $8\, s$. The speed of $B$ is (in $km/h$):

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