Question

A boat leaves a dock at 5:00 pm and travels due south at a speed of 20 km/h. another boat has been heading due east at 15 km/h and reaches the same dock at 6:00 pm. how many minutes after 5:00 pm were the two boats closest together? (round your answer to the nearest minute.

Answer 1

Place the dock at (0, 0) in the xy-plane. At 5:00 P.M. boat A is at (0, 0). It's position after 5:00 P.M. is given by (0, -20t) where t is in hours. At 6:00 P.M. boat B is at (0, 0). That's 1 hour after boat A left the point (0, 0) so 1 h x 15 Km/h=15 Km which means at 5:00 P.M. boat B was 15 Km west of the dock at (0, 0) which means it was at (-15, 0) at 5:00 P.M. Boat B's position after 5:00 P.M. is therefore (-15+15t, 0). Use the distance formula to find the distance between the two boats. 
d=√((x2-x1)²+(y2-y1)²) 
=√((-15+15t-0)²+(0+20t)²) 
=√(225-450t+225t²+400t²) 
=√(225-450t+625t²) 
Find the derivative 
d'= (1/2)(225-450t+625t²)^(-1/2)(-450+1250t) 
Set equal to zero and you get 
-450+1250t=0 
t=450/1250 
=0.36 h=22 minutes 

Answer 2

The time taken by the two boats to be closest to each other is $\boxed{22{\text{ minutes}}}$

Further explanation:

The formula for speed is given as,

$\boxed{s=\dfrac{d}{t}}$

Here, $s$ represents the speed, $d$ represents the distance, and $t$ represents time.

The distance between the two points $\left( {{x_1},{y_1}}\right)$ and $\left( {{x_2},{y_2}}\right)$ can be expressed as follows,

$\boxed{D=\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} }$

The D is distance between the two points.

Given:

A boat leaves a dock at 5:00 pm and travels due south at a speed of 20 km/h. another boat has been heading due east at 15 km/h and reaches the same dock at 6:00 pm.

Calculation:

Consider the dock is at $\left( {0,0}\right)$ in two dimensions that is $xy - {\text{plane}}$.

Boat A is at $\left( {0,0}\right)$ at 5 P.M. The distance after 5 P.M can be calculated as follows,

$\begin{aligned}- 20=\frac{d}{t}\\- 20t=d\\\end{aligned}$

The d is the distance and t is the time. -20 represents that the boat is moving in the south direction with speed 20 km/h.

The position of the boat will be $\left({0, - 20t}\right)$.

The boat B was 15 km west of the dock at $\left( {0,0}\right)$ which means the boat is at $\left( { - 15,0}\right)$ at 5 PM.

Boat B is at $\left( {0,0}\right)$ at 6 P.M. The distance after 6 P.M can be calculated as follows,

$\begin{aligned}15&=\frac{d}{t}\\15t&=d\\\end{aligned}$

The d is the distance and t is the time. 15 represents that the boat is moving in the east direction with speed 15 km/h.

The position of the boat will be $\left( { - 15 + 15t,0}\right)$.

The distance between the boats can be calculated as follows,

$\begin{aligned}D&=\sqrt {{{\left({ - 15 + 15t - 0}\right)}^2} + {{\left( {0 + 20t}\right)}^2}}\\&= \sqrt {{{\left({ - 15 + 15t} \right)}^2} + {{\left( {20t} \right)}^2}}\\&=\sqrt {225 - 450t + 225{t^2} + 400{t^2}}\\&=\sqrt {225 - 450t + 625{t^2}}\\\end{aligned}$

Differentiate the above equation with respect to t.

$\dfrac{{dD}}{{dt}}=\dfrac{1}{2} \times {\left( {225 - 450 + 625{t^2}} \right)^{-\dfrac{1}{2}}}\times\left({ - 450 + 1250t}\right)$

Now substitute 0 for $\dfrac{{dD}}{{dt}}$ to obtain the time.

$\begin{aligned}\dfrac{1}{2}\times{\left({225 - 450 + 625{t^2}}\right)^{ - \dfrac{1}{2}}} \times \left( { -450 + 1250t} \right)7= 0\\\left( { - 450 + 1250t} \right)&=0\\t&=\frac{{450}}{{1250}}\\&=0.36{\text{ hours}}\\\end{aligned}$

The time in minutes will be $t = 22{\text{ minutes}}$.

The time taken by the two boats to be closest to each other is $\boxed{22{\text{ minutes}}}$.

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Linear equation

Keywords: Boat, leaves, dock, travels, south, speed, time, heading, east, reaches, 20km/h, two boats, closest, north, west, minutes.