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​ (pls answer asap)​

Answer 1

Answer:

Explanation: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

✪ANSWER✪

• They are closest at 57.6 minute pass noon.

• Shortest distance between them is 18 Km

★EXPLAINATION★

☆ɢɪᴠᴇɴ☆

• Ship A, sailing at v₁ = 15 Km/ hr, East passes a certain point(say Origin) at noon.

• Ship B, sailing at v₂ = 20 Km/hr North passes through the same point at 1. 30 PM.

☆ᴛᴏ ғɪɴᴅ☆

• Time of closest distance between A and B.

• The closest distance between A and B.

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

Let us assume that

• The certain point is the origin.

• Our observation starts at noon, so at noon t = 0.

᯽EXPRESSION OF DISTANCES

Let x(t), y(t) be the distance of ship A and B from origin and D(t) be the distance between the ships. Using equation of motion

$v₁ = \frac{x(t) - x(0)}{t - 0}$

$⇒x(t) - x(0) = v₁t$

$⇒x(t) - 0 = 15t$

$⇒x(t) = 15t Km$

$v₂ = \frac{y(t) - y(1.5)}{t - 1.5}$

$⇒y(t) - x(1.5) = v₂(t - 1.5)$

$⇒y(t) - 0 = 20(t- 1.5)$

$⇒y(t) = 20(t- 1.5) Km$

Since A and B are moving along perpendicular directions,

$D(t)² = x(t)² + y(t)²$

᯽TIME OF SHORTEST DISTANCE

To calculate time of shortest distance, let us calculate D′(t) and equate it to zero. Differenting above equation with time, we get

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

$⇒0 = 15t\times 15 + 20(t- 1.5)\times 20$

$⇒(225 + 400)t = 400\times 1.5$

$⇒ t = \frac{600}{625}$

$⇒ t = \frac{24}{25}\:\: Hr$

$⇒ t = \frac{24}{25}\times 60\:\: min$

$⇒ t = 57.6\:\: min$

Therefore, the ship become closest at 57.6 min pass noon.

᯽SHORTEST DISTANCE

Substituting above value of t in D(t) will give the shortest distance.

$D² = x(t=\frac{24}{25})² + y(t=\frac{24}{25})²$

$⇒D² = (15\times \frac{24}{25})² + (20(\frac{24}{25}-1))²$

$⇒D² = (14.4)² + (19.2 - 30)²$

$⇒D² = 324$

$⇒D = 18\:\:Km$

I hope you understand this solution.There is another method to solve this question but the solution was getting quite lengthy. So I posted just one method.