By cosine law,

$s^2 = 20^2 + (10t)^2 - 2(20)(10t) \cos 60^\circ$

$s = \sqrt{100t^2 - 200t + 400}$

$\dfrac{ds}{dt} = \dfrac{200t - 200}{2\sqrt{100t^2 - 200t + 400}}$

$\dfrac{ds}{dt} = \dfrac{100t - 100}{\sqrt{100t^2 - 200t + 400}}$

after 6 hrs from start, t = 6 - 2 = 4 hrs

$\dfrac{ds}{dt} = \dfrac{100(4) - 100}{\sqrt{100(4^2) - 200(4) + 400}}$

$\dfrac{ds}{dt} = \dfrac{100(4) - 100}{\sqrt{100(4^2) - 200(4) + 400}}$

$\dfrac{ds}{dt} = 8.66 \, \text{ mi/hr}$ *answer*

## Comments

## Dear Sir,

Dear Sir,

Reference to problem no. 37 , Why the time 6 hours deducted by 2 hours..

Thanks .