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Question

(A) Let the distance be $d \, km$ and the initial speed be $s \, km/h$.From the first condition: $\frac{d}{s} - \frac{d}{s+6} = 4 \implies \frac{6d}{s

Vedclass · Accepted Answer

$720$

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(A) Let the distance be $d \, km$ and the initial speed be $s \, km/h$.From the first condition: $\frac{d}{s} - \frac{d}{s+6} = 4 \implies \frac{6d}{s(s+6)} = 4 \implies \frac{d}{s(s+6)} = \frac{2}{3} \implies d = \frac{2s(s+6)}{3} \dots (i)$From the second condition: $\frac{d}{s-6} - \frac{d}{s+6} = 10 \implies d \left( \frac{(s+6) - (s-6)}{(s-6)(s+6)} ight) = 10 \implies d \left( \frac{12}{s^2-36} ight) = 10 \implies d = \frac{10(s^2-36)}{12} = \frac{5(s^2-36)}{6} \dots (ii)$Equating $(i)$ and $(ii)$: $\frac{2s^2+12s}{3} = \frac{5s^2-180}{6} \implies 4s^2 + 24s = 5s^2 - 180 \implies s^2 - 24s - 180 = 0$.Solving the quadratic equation: $(s-30)(s+6) = 0$. Since speed cannot be negative,$s = 30 \, km/h$.Substituting $s = 30$ into $(i)$: $d = \frac{2(30)(36)}{3} = 20 	imes 36 = 720 \, km$.

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