question_answer

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 \[x=\frac{d}{\sqrt{{{2}^{2}}-1}}=\frac{d}{\sqrt{3}}\]