**If she walks at $3\, mi/hr$ and rows at $2\, mi/hr$, at which point on the shore should she land to minimize the total travel time?****If she walks at $3\, mi/hr$, what is the minimum speed at which she must row so that the quickest way to the restaurant is to row directly (with no walking)?**

The purpose of this math question is to find the minimum travel time and minimum distance.

One of the most major aspects of Classical Mechanics is the phenomenon of motion in physics. The moving of an object is the change in its location relative to a fixed point. Similarly, the change in position of an object relative to its surroundings in a given period is referred to as motion. Distance, displacement, speed, velocity, time, and acceleration are the terms to characterize the motion of an object having mass. An object is considered to be at rest, immobile, motionless, static, or to possess a fixed or time-independent position with respect to its surroundings if it does not change relative to a given reference frame.

Distance is defined as an object’s net movement without any direction. Distance and displacement are two measures that appear to have the same meaning but have very distinct meanings and definitions. Distance is defined as “how much surface area is covered throughout an object’s motion,” whereas displacement is defined as “how far from the place an object is.” Distance is a scalar attribute, which means this only refers to the entire magnitude and doesn’t take into consideration start or endpoints.

## Expert Answer

Let $x$ represent the distance between the closest point on a shoreline and where the woman lands. This implies that the distance between where she lands and the restaurant is $(6 – x)\,mi$.

Let $t$ be the amount of time it takes her to reach the restaurant. To perform this minimization, write $t$ as a function of $x$ and then equate its derivative to $0$.

Now, using the Pythagoras theorem, the distance between the boat and the point where the woman lands is:

$d=\sqrt{4^2+x^2}$

$d=\sqrt{16+x^2}$

Also, the time is:

$t(x)=\left(\dfrac{\sqrt{16+x^2}}{2}-\dfrac{6-x}{3}\right)\,hr$

$\dfrac{dt}{dx}=\dfrac{2x}{4\sqrt{16+x^2}}-\dfrac{1}{3}$

$\dfrac{dt}{dx}=\dfrac{x}{2\sqrt{16+x^2}}-\dfrac{1}{3}$

Now, for the minimum time:

$\dfrac{dt}{dx}=0$

$\dfrac{x}{2\sqrt{16+x^2}}-\dfrac{1}{3}=0$

$3x=2\sqrt{16+x^2}$

$9x^2=4(16+x^2)$

$5x^2=64$

$x=\pm\,\dfrac{8}{\sqrt{5}}\,mi$

Since the distance is always positive, so $x=\dfrac{8}{\sqrt{5}}\,mi=\dfrac{8\sqrt{5}}{5}\,mi$.

Now, if the woman lands at a point which is $6\,mi-\dfrac{8\sqrt{5}}{5}\,mi=\dfrac{30-8\sqrt{5}}{5}\,mi$ far from the restaurant, she will minimize the time it takes to reach the restaurant.

## Example

Two women begin walking a certain distance at the same time, one at $5\, kmph$ and the other at $4\, kmph$. The former arrives one hour before the latter does. Determine the distance.

### Solution

Let $x\,km$ be the distance required, then:

$\dfrac{x}{4}-\dfrac{x}{5}=1$

$\dfrac{5x-4x}{20}=1$

$x=20\,km$