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.

A ship is moving at a constant speed of 10 km//hr in the direction of

1.	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» Solution :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.