Question

A dispatch rider is in open country at a distance of 6 km from the nearest point P of a straight road. He wishes to proceed as quickly as possible to a point Q on the road 20 km from P. If his maximum speed across country is 40 km/hr, then at what distance from P, he should strike the road at the speed 50 km/hr.

• A. 8 km
• B. 6 km
• C. 7 km
• D. ) None of these

Answer 1

Answer: (A)

8 km

Answer 2

Let A be the initial position of rider.

Let, PB = x

\ QB = 20 – x

and AB = $\sqrt { \mathrm { AP } ^ { 2 } + \mathrm { BP } ^ { 2 } }$

= $\sqrt { 6 ^ { 2 } + x ^ { 2 } }$ km.

\ Total time T,

T = + $\frac { 20 - x } { 50 }$

= $\frac { 1 } { 40 }$. – $\frac { 1 } { 50 }$

For maximum and minimum value, must have

= 0

Ž = $\frac { 1 } { 50 }$

or =

or – x² = 36

Žx² = $\frac { 36 \times 16 } { 9 }$

Ž x = $\frac { 6 \times 4 } { 3 }$ = 8 km.