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Class 11

Physics

Motion in a Plane

Relative motion

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Question

Hard

Solving time: 4 mins

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Book:

Problems in Physics I for IIT JEE (McGraw Hill)

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Text solutionVerified

Let the man travel a distance '

x

' on road in order to reach '

P

' in minimum time
Solution Image

t = \frac{x}{v _{1}} + \frac{d ^{2} + ( l - x ) ^{2}}{v _{2}}

....(A)
Time to travel from

A

P

depends on

x

. This time

(t)

is minimum when

\frac{d t}{d x} = 0

\Rightarrow \frac{1}{v _{1}} + \frac{1}{v _{2}} \frac{ℓ - x}{d ^{2} + ( ℓ - x ) ^{2}} = 0

Solving,

x = l \pm \frac{v _{2} d}{v _{1}^{2} - v _{2}^{2}}

The positive sign in unacceptable as

x

cannot be larger than

l

∴ x = l - \frac{v _{2} d}{v _{1}^{2} - v _{2}^{2}}

(a) If

\frac{v _{2} d}{v _{1}^{2} - v _{2}^{2}} \geq l

\Rightarrow d \geq ℓ \frac{v _{1}^{2} - v _{2}^{2}}{v _{2}} = 50 \times \frac{4}{3} = \frac{200}{3} m

In this case

x

is negative or zero.
But it makes no sense to have

x < 0

.
Hence, man will need to walk along straight line

A P

in order to cover the distance in minimum time.
(b)

∴

l > \frac{v _{2} d}{v _{1}^{2} - v _{2}^{2}}

then

x

is positive.

x = l - \frac{v _{2} d}{v _{1}^{2} - v _{2}^{2}} = 50 - \frac{3 \times \frac{100}{3}}{5 ^{2} - 3 ^{2}} = 50 - 25 = 25 m

Putting this in equation (A)

t_{m i n} = \frac{25}{5} + \frac{( \frac{100}{3} ) ^{2} + ( 50 - 25 ) ^{2}}{3} = \frac{170}{9} s

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Question Text
Topic	Motion in a Plane
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	57

Text solutionVerified

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