1.

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.

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`.


Discussion

No Comment Found