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Two men A and B start with velocities v at the same time from the junction of two roads inclined at `45^@` to each other.If they travel by different roads,find the rate at which they are being separated. |
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Answer» With the given details, we can create a right angle triangle. Please refer to video for the figure. From the figure, `(OP)/(OQ) = tan45^@ = 1` `:. OP = OQ` Let `OP = OQ = k` Then, `PQ = sqrt(k^2+k^2) = ksqrt2` Now, in right angle triangle `OPQ`, `OP^2+OQ^2 = PQ^2` As we have to find the rate of separation, we will differentiate above equation. So, `2OP(d(OP))/dt+2OQ(d(OQ))/dt = 2PQ(d(PQ))/dt` As we are given velocities of two person as `v`, `:. (d(OP))/dt = (d(PQ))/dt = v` So, our equation becomes, `=>2kv+2k(d(OQ))/dt = 2sqrt2kv` `=>v+(d(OQ))/dt =sqrt2v` `=> (d(OQ))/dt = (sqrt2-1)v` So, rate of separation will be `(sqrt2-1)v`. |
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