Kinematics-3

1 Answers

2305

Shaswata Roy ·2013-04-06 02:49:33

(i)t(x)=\frac{\sqrt{9^{2}+x^{2}}}{4}+\frac{15-x}{5}

(ii)\frac{\mathrm{d}t(x)}{\mathrm{d}x}=\frac{x}{4\sqrt{9^{2}+x^{2}}}-\frac{1}{5}

t is minimum when dtdx=0 and d t²d² x>0

Therefore x = 12km

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