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HETAV PATEL

Grade 11,

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A man in a car at location Q on a straight highway is moving with speed V . He decide to reach point P in a field at a distance d from the heighway(point M) as shown in figure . The speed of car in the field is half to that on the heighway .What should be the distance RM so that the time taken to reach P is minimum .

A man in a car at location Q on a straight highway is moving with speed V . He decide to reach point P in a field at a distance d from the heighway(point M) as shown in figure . The speed of car in the field is half to that on the heighway .What should be the distance RM so that the time taken to reach P is minimum .

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2 Answers

Arun
25750 Points
5 years ago

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}}

 

Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Given speed of the vehicle in highway = V m/s
and its speed in field =2v​ m/s
Time taken to travel from R to P = ​RP/Vinfield ​= 2*sqare root of ((z−x)^2+d^2)​​/v
The value of RP is found by using Pythagoras theorem.
Now total time taken to travel from point Q to point P taking diversion from R
t total ​= x/v ​+ 2*sqare root of ((z−x)^2+d^2)​​/v
For the time T to minimum,
dT/dx ​= 0
1/v ​+ (2/v)*(1/(2*sqare root of ((z−x)^2+d^2)​​))​*(−2)*(z−x) = 0
(−2)(z−x) = sqare root of ((z−x)^2+d^2)​​
​4(z−x)^2 = d^2 + (z−x)^2
z−x = d/ sqare root of (3)

Thanks and Regards

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