question_answer

OA and OB are two roads enclosing an angle of\[{{120}^{o}}\]. X and y start from ?O' at the same time. X travels along OA with a speed of 4 km/h and Y travels along OB with a speed of 3 km/h. The rate at which the shortest distance between X and V is increasing after \[1\text{ }h\]is

A)  \[\sqrt{37}km/h\]

B)  \[37m/h\]

C)  \[13km/h\]     

D)  \[\sqrt{13}km/h\]

Correct Answer: A

Solution :

Given, speed of x is \[4\text{ }km/h\]and y is \[\text{3 }km/h\]. After time r the distance covered by x is 41 and y is 3t.                                 Let shortest distance between x and \[y=A\]. Then by cosine law \[{{A}^{2}}={{(4t)}^{2}}+{{(3t)}^{2}}-(4t)\,(3t)\,2\cos {{120}^{o}}\] \[\Rightarrow \] \[{{A}^{2}}=16{{t}^{2}}+9{{t}^{2}}-24{{t}^{2}}\left( -\frac{1}{2} \right)\] \[\Rightarrow \]  \[{{A}^{2}}=25{{t}^{2}}+12{{t}^{2}}\] \[\Rightarrow \] \[{{A}^{2}}=37{{t}^{2}}\]                      ??(i) \[\Rightarrow \]\[A=\sqrt{37}t\] If  \[t=1h,\] then \[A=\sqrt{37}km\] Now, differentiating Eq. (i) w.r.t. t, we get                 \[2AA'=37(2t)\] After  \[t=1\text{ }h,\]we get                 \[2\sqrt{37}A'=2(37)\] \[\Rightarrow \]               \[A'=\sqrt{37}\] Thus, rate at which shortest distance A changes with time is \[\sqrt{37}km/h\].