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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 the 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 ?

  • Option 1)

    d

  • Option 2)

    \frac{d}{\sqrt{2}}

  • Option 3)

    \frac{d}{2}

  • Option 4)

    \frac{d}{\sqrt{3}}

 

Answers (2)

As we have learnt

Average Speed -

Amount of total distance covered in total time.

Formula

Average\, speed =\frac{total \, distance\, covered}{total \, time\, taken}

V_{av}= \frac{s}{t}

 

- wherein

E.g A body covers a total distance of 50 m with variable speed in 5 sec. Find average speed of body during this time
interval.

V_{av}= \frac{s}{t}

\Rightarrow \frac{50m}{5s}= 10m/s

Average Speed = 10 m/s

 

 

 Time taken to travel from Q to R =\frac{x}{v}

Time taken to travel from R to P =\frac{\sqrt{d^{2}+\left ( l-x \right )^{2}}}{v/2}

Total time t =\frac{x}{v}+\frac{2.\sqrt{d^{2}+\left ( l-x \right )^{2}}}{v}

For t to be min,\frac{\mathrm{d} T}{\mathrm{d} x}=0

\therefore \frac{1}{v}+\frac{2}{v}.\frac{1}{2\sqrt{d^{2}+\left ( L-x \right )^{2}}}.-2\left ( L-x \right )=0 or

\therefore 2\left ( L-x \right )= {\sqrt{d^{2}+\left ( L-x \right )^{2} or 4\left ( L-x \right )^{2} = d^{2}+\left ( L-x \right )^{2}

L-x = \frac{d}{\sqrt{3}}

 


Option 1)

d

Option 2)

\frac{d}{\sqrt{2}}

Option 3)

\frac{d}{2}

Option 4)

\frac{d}{\sqrt{3}}

Posted by

Vakul

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