Question

Ship A is sailing towards the northeast with velocity `vecv = 30hati + 50hatj` km/hr where `hati` points east and `hatj`, north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing west at 10 km/hr. A will be at the minimum distance from B in ______.

पर्याय

• 4.2 hrs.

• 2.6 hrs.

• 3.2 hrs.

• 2.2 hrs.

Answer 1

Ship A is sailing towards the northeast with velocity `vecv = 30hati + 50hatj` km/hr where `hati` points east and `hatj`, north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing west at 10 km/hr. A will be at the minimum distance from B in 2.6 hrs.

Explanation:

Given: the speed of ship A,

`vecv_A = (30hati + 50hatj)` km/h,

the B ship's speed,

`vecv_B = -10hati` km/h

The distance between the two ships

= `vecr_B - vecr_A = 80hati + 150hatj`.

To find: The time when the distance between the two ships is shortest.

Let `vecr_A = 0hati + 0hatj;` that gives:

`vecr_B = 80hati + 150hatj`

After time t:

`vecr_A = 30 t hati + 50t;`

Ship B is sailing to the west:

`vecr_B = (80 - 10t)hati + 150hatj`

After time t, the distance between the two ships is:

`d = vecr_B - vecr_A`

= `(80 - 10t - 30t)hati + (150 - 50t)hatj`

`d^2 = (80 - 40t)^2 + (150 - 50t)^2`

When the gap between the two ships is at its smallest, t, then:

`d/dt(d^2) = 0`

`d/dt[(80 - 40t)^2 + (150 - 50t)^2] = 0`

2(80 - 40t)(-40) + 2(150 - 50t)(-50) = 0

-3200 + 1600t - 7500 + 2500t = 0;

t = `107/41;`

t = 2.6