Mathematics, 05.11.2020 19:20 JvGaming2001

Brandon is on one side of a river that is 70 m wide and wants to reach a point 250 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way. Find the minimum amount of time if Brandon can swim at 2 m/s and run at 6 m/s.

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Answer from: ic838847

Minimum amount of time to reach the desired point is 79.5 s

Brandon is on one side of a river

The width of the river = 70 m

Brandon wants to reach a point that is 250 m downstream of opposite side as quickly as possible by swimming diagonally across the river then running the rest of the way.

The swimming speed of Brandon = 2 m/s

The speed of running of Brandon = 6 m/s

Let us consider that the included angle be 45°

$\rm cos \; \theta = \dfrac{70}{x} \\\\cos \; 45 = \dfrac{70}{x} \\\x = \dfrac{70}{0.707} \approx 99$

So the distance traveled by Brandon in water is 99 m

As the swimming speed of Brandon is given 2 m/s

So the time taken by Brandon to cross the river is given as follows

$\rm Time = \dfrac{Distance}{Speed}$

$\rm Time \; to\; cross\; the\; river = \dfrac{99}{2} = 49.5 \; s$

Let the horizontal distance covered by Brandon by swimming is H

So from Pythagorean theorem we can write

$\rm H = \sqrt{99^2 -70^2 } = 70$

The distance that is left to be traveled by running is 250 - 70 = 180 m

So Time taken in running to reach the point by running = 180/6 = 30 s

So the time to reach the desired point = 49.5 + 30 = 79.5 s

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Answer from: datbih12345

The minimum amount of time required is 74.7 seconds.

Step-by-step explanation:

The question can be solved in two cases:

i. diagonal swimming of Brandon across the river

ii. Running of the rest distance by Brandon.

Case I: Since he would swim diagonally across the river, then the diagonal distance can be determined by applying appropriate trigonometric function.

Let the diagonal be represented by x, an included angle θ = $45^{o}$ is formed at his starting point (angle formed by the diagonal and the length of the width of the river).

So that:

Cos θ = $\frac{adjacent}{hypotenuse}$

Cos $45^{o}$ = $\frac{70}{x}$

x = $\frac{70}{Cos 45^{o} }$

= 98.995

x = 99 m

The diagonal distance covered by Brandon is 99 m.

The time to cover this distance can be calculated by:

speed = $\frac{distance}{time}$

time = $\frac{distance}{speed}$

Given that his swimming rate is 2 m/s, then:

time = $\frac{99}{2}$

= 49.5

The time taken by Brandon to swim diagonally is 49.5 seconds.

Case II: Since he want to cover a total distance of 250 m;

remaining distance to be covered = 250 - 99

= 151

time = $\frac{distance}{speed}$

Given that he runs at 6 m/s, then;

time = $\frac{151}{6}$

= 25.1667

The time taken by Brandon to run the remaining distance is 25.2 seconds.

The minimum amount of time required = time taken by Brandon to swim diagonally + time taken by Brandon to run the remaining distance

= 49.5 + 25.2

= 74.7

The minimum amount of time required is 74.7 seconds.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

$x=-2$ or $x=-5$

step-by-step explanation:

we need to solve for 'x' the following equation: $\sqrt{(x+6)} - 4 = x$

we have that:

$\sqrt{(x+6)} - 4 = x$ ⇒ $(x+6) = (x+4)^{2}$

⇒ $x+6 = x^{2} + 8x + 16$ ⇒ $x^{2} + 7x + 10 = 0$

factorizing, we get that:

$x^{2} + 7x + 10 = 0$ ⇒ $(x+2)(x+5)=0$

then, the solutions are: $x=-2$ or $x=-5$

nothing extraneous has been found!

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