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 \[\sqrt{37}km/h\].You need to login to perform this action.
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