A man in a car at location \[Q\] on a straight highway is moving with speed\[v\]. He decides to reach a point \[P\] in a field at a distance \[d\] from highway (point\[M\]) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance\[RM\], so that the time taken to reach \[P\] is minimum? |
[JEE Online 15-04-2018 (II)] |
A) \[\frac{d}{\sqrt{3}}\]
B) \[\frac{d}{2}\]
C) \[\frac{d}{\sqrt{2}}\]
D) \[d\]
Correct Answer: B
Solution :
[b] Let the car turn of the highway at a distance 'x' from the point |
\[M.So,\] \[RM=x\] |
And if speed of car in field is \[v\], then time taken by the car to cover the distance \[QR=QM-x\]on the highway, \[{{t}_{1}}=\frac{QM-x}{2v}.....(1)\] |
the time taken to travel the distance 'RP' in the field |
\[{{t}_{2}}=\frac{\sqrt{{{d}^{2}}+{{x}^{2}}}}{v}\]...(2) |
So, the total time elapsed to move the car from\[Q\] to \[P\] |
\[t={{t}_{1}}+{{t}_{2}}=\frac{QM-x}{2v}+\frac{\sqrt{{{d}^{2}}+{{x}^{2}}}}{v}\] |
for \['t'\] to be minimum |
\[\frac{dt}{dx}=0\] |
\[\frac{1}{v}\left[ -\frac{1}{2}+\frac{x}{\sqrt{{{d}^{2}}+{{x}^{2}}}} \right]=0\] |
Or \[x=\frac{d}{\sqrt{{{2}^{2}}-1}}=\frac{d}{\sqrt{3}}\] |
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