Question

at noon ship a is sailing east with a velocity of 15 km/hr it passes a certain point. a second ship sailing north with a speed of 20 km/hr passes the same point at 1:30 pm at what time they are closest together and what is the shortest distance between them​

Answer 1

Let x = the distance of ship A west of the north-south line through B.

y = the distance of ship B north of the east-west line through A.

At any time the distance apart is given by L^2 = x^2 + y^2.

Then:

2L dL/dt = 2x dx/dt + 2y dy/dt

L dL/dt = x(-35) + y(25)

dL/dt = [-35x + 25y]/L

At 4:00 p.m.:

x = 150-140 = 10

y = 100

L^2 = 100+10000 = 10100 => L = 100.4988

So:

dL/dt = [-350 + 2500]/100.4988 = 21.393 km/hr

So at 4:00 p.m., the ships are separating at a speed of 21.393 km/hr.

Answer 2

Concept:

When an object is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.

Given:

A ship is sailing east with a velocity of 15 km/hr at noon.

A second ship is sailing north with a velocity of 20 km/hr and passes the point at 1:30 pm.

Find:

The time at which they are closest to each other and the shortest distance between them.

Solution:

The speed of ship A, $v_1=15km/h$ passes a certain point at noon.

The speed of ship B, $v_2=20km/h$ passes the same point at 1:30 pm.

Initially, the time is $t=0$ as observation starts at the point for ship A.

Let $x(t),y(t)$ be the distance of ships A and B from the point and D(t) be the distance between the two ships.

By equation of motion,

$v=\frac{x(t)-x(0)}{t-0}$

$vt=x(t)\\x(t)=15t$

When ship B passes the point at 1:30 pm i.e 1.5 hr from noon.

$v_2=\frac{y(t)-y(1.5)}{1-1.5}$

$20(t-1.5)=y(t)-0\\y(t)=20(t-1.5)$

Now ship A is moving east and ship B is moving north. Therefore, both ships are moving in perpendicular directions.

The distance between the ships:

$D^2(t)=x^2(t)+y^2(t)$

Differentiating the equation with respect to time,

$D'(t)=2x(t)x'(t)+2y(t)y'(t)\\D'(t)=2x(t)v_1+2y(t)v_2$

The distance is shortest when $D'(t)=0$

$2x(t)v_1+2y(t)v_2=0\\15t\times 15+20(t-1.5)20=0\\t=57.6min$

$t=\frac{24}{25}hr$

Now,

$D^2=x^2(t)+y^2(t)\\D^2=(15\times \frac{24}{25})^2+(20\times \frac{24}{25})^2$

$D^2=(14.4)^2+(-10.8)^2\\D^2=324\\D=18km$

The time when the ship is closest at $57.6min$ from noon and the shortest distance between the ships is $18km$.

#SPJ3