A ship is moving at a constant speed of `10 km//hr` in the direction of the unit vector `hat(i)`. Initially. Its position vector, relative to a fixed origin is `10(-hat(i)+hat(j))` where `hat(i)` & `hat(j)` are perpendicular vectors of length 1 km. Find its position vector relative to the origin at time t hours later. A second ship is moving with constant speed `u km//hr` parallele to the vector `hat(i)+2hat(j)` and is initially at the origin
(i) If `u=10 sqrt(5) km//h`. Find the minimum distance between the ships and the corresponding value of t
(ii) Find the value of u for which the shipd are on a collision course and determine the value of t at which the collision would occur if no avoiding action were taken.

Answer

`vec(s)=10(t-1)hat(i)+10hat(j)(a) t=1/2 h, 10 km` (b) `(10sqrt(5))/2, t=2 hr`

Explanation

It's velocity is `10 hat (i)`

`:.` displacement after time `'t' = 10 hat (i) xx t`
Velocity of second ship `= u xx ((hat (i) +2 hat(j)))/(sqrt(5))`
`tan theta = (2u//5)/((10 -(u)/(sqrt(5))))=(2 xx 10 sqrt(5))/(10 sqrt(5) -10 sqrt(5))`
(i) `t = (10)/(20) = (1)/(20) sec`, minimum distance = 10 km.