A man in a car at location Q on a straight highway is moving with speed \upsilon . 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 ?

Motion in A Plane PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q56 with Solution

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Correct answer: D

Interpret the figure and define variables

Let the car move from point $Q$ along the highway to some point $R$ , and then leave the highway through the field to reach point $P$ .

Let:

$RM = x$
$PM = d$ (perpendicular distance of $P$ from the highway)
speed on highway $= \upsilon$
speed in field $= \dfrac{\upsilon}{2}$

Since only the variable part matters for minimization, we minimize the time from $R$ to $P$ together with the highway segment up to $R$ .

Write the total time as a function of $x$

If the car goes from $Q$ to $R$ on the highway, and then from $R$ to $P$ through the field, then:

Highway distance contributes time proportional to $QR/\upsilon$
Field distance is

RP = \sqrt{x^2 + d^2}

so field time is

\frac{RP}{\upsilon/2} = \frac{2\sqrt{x^2+d^2}}{\upsilon}

Now, as $R$ moves toward $M$ by distance $x$ , the highway distance from $Q$ changes linearly. From the figure setup, minimizing total time is equivalent to minimizing

T(x)= -\frac{x}{\upsilon} + \frac{2\sqrt{x^2+d^2}}{\upsilon}

(the constant part independent of $x$ is omitted).

So we minimize

f(x) = -x + 2\sqrt{x^2+d^2}

Differentiate and set equal to zero

\frac{df}{dx} = -1 + 2\cdot \frac{x}{\sqrt{x^2+d^2}}

For minimum time,

-1 + \frac{2x}{\sqrt{x^2+d^2}} = 0

So,

\frac{2x}{\sqrt{x^2+d^2}} = 1

Squaring both sides,

4x^2 = x^2 + d^2

3x^2 = d^2

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

Thus,

RM = \frac{d}{\sqrt{3}}

Check the options

A: $d$ ❌
B: $\dfrac{d}{\sqrt{2}}$ ❌
C: $\dfrac{d}{2}$ ❌
D: $\dfrac{d}{\sqrt{3}}$ ✅

Hence the correct option is D.

More from Motion in A Plane

View written solutionFree

Correct answer: D

Interpret the figure and define variables

Let the car move from point $Q$ along the highway to some point $R$ , and then leave the highway through the field to reach point $P$ .

Let:

$RM = x$
$PM = d$ (perpendicular distance of $P$ from the highway)
speed on highway $= \upsilon$
speed in field $= \dfrac{\upsilon}{2}$

Since only the variable part matters for minimization, we minimize the time from $R$ to $P$ together with the highway segment up to $R$ .

Write the total time as a function of $x$

If the car goes from $Q$ to $R$ on the highway, and then from $R$ to $P$ through the field, then:

Highway distance contributes time proportional to $QR/\upsilon$
Field distance is

RP = \sqrt{x^2 + d^2}

so field time is

\frac{RP}{\upsilon/2} = \frac{2\sqrt{x^2+d^2}}{\upsilon}

Now, as $R$ moves toward $M$ by distance $x$ , the highway distance from $Q$ changes linearly. From the figure setup, minimizing total time is equivalent to minimizing

T(x)= -\frac{x}{\upsilon} + \frac{2\sqrt{x^2+d^2}}{\upsilon}

(the constant part independent of $x$ is omitted).

So we minimize

f(x) = -x + 2\sqrt{x^2+d^2}

Differentiate and set equal to zero

\frac{df}{dx} = -1 + 2\cdot \frac{x}{\sqrt{x^2+d^2}}

For minimum time,

-1 + \frac{2x}{\sqrt{x^2+d^2}} = 0

So,

\frac{2x}{\sqrt{x^2+d^2}} = 1

Squaring both sides,

4x^2 = x^2 + d^2

3x^2 = d^2

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

Thus,

RM = \frac{d}{\sqrt{3}}

Check the options

A: $d$ ❌
B: $\dfrac{d}{\sqrt{2}}$ ❌
C: $\dfrac{d}{2}$ ❌
D: $\dfrac{d}{\sqrt{3}}$ ✅

Hence the correct option is D.

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