Question 125: Motion in 2D & Vectors

Question

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 ?

Accepted Answer

Let the distance QM = l and distance RM = x.
Time to reach from Q to R is $${t_1} = {{l - x} \over v}$$
Time to reach from R to P is $${t_2} = {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$$
Therefore, $$t = {t_1} + {t_2} = {{l - x} \over v} + {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$$
On differentiating, we get
$${{dt} \over {dx}} = {{0 - 1} \over v} + {1 \over {v/2}}{1 \over {2\sqrt {{x^2} + {d^2}} }} 	imes 2x$$
$$ \Rightarrow {{dt} \over {dx}} = {{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }}$$
Put $${{dt} \over {dx}} = 0$$, we get
$${{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = 0$$
$$ \Rightarrow {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = {1 \over v}$$
$$ \Rightarrow 2x = \sqrt {{x^2} + {d^2}}  \Rightarrow 4{x^2} = {x^2} + {d^2}$$
$$ \Rightarrow 3{x^2} = {d^2} \Rightarrow x = {d \over {\sqrt 3 }}$$
Therefore, the distance $$RM = x = {d \over {\sqrt 3 }}$$, the time taken to reach P is minimum.

Answer

d

Answer

$${d \over {\sqrt 2 }}$$

Answer

$${d \over 2}$$

Answer

$${d \over {\sqrt 3 }}$$

Motion in 2D & Vectors

Choose the correct answer: