A and B take part in a 100m race. A runs at 5km/hr. A gives B a start of 8m and still beats him by 8 sec. B speed is:
Answer: D
Find A's time first.
\[ 5 \text{ km/hr} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]
\[ t_{A} = \frac{100}{25/18} = 72 \text{ s} \]
B gets an 8 m start, so B runs 92 m, and finishes 8 seconds after A:
\[ t_{B} = 72 + 8 = 80 \text{ s} \]
\[ v_{B} = \frac{92}{80} = 1.15 \text{ m/s} = 1.15 \times \frac{18}{5} = 4.14 \text{ km/hr} \]
The start reduces B's distance to 92 m; the 8 seconds are added to A's time.
Please log in or register to report this question.