A boat A moving with speed 8kmh leaves point p whose position vector is r1=4i+3j heading directly towards point Q whose position vector is r2=2i+5j. At the same time another boat B leaves point 0i+0j. if the velocity of boat B is vector Vba=root2(-5i+3j), find

a. the veloctiy of boat B
b. the timetaken when the boats are close
c. the shortest distance apart

To solve this problem, we need to find the velocity of boat B, the time taken when the boats are close, and the shortest distance between the two boats.

a. Velocity of boat B:
The velocity of boat B is given as vector Vba = √2(-5i + 3j).

b. Time taken when the boats are close:
We can calculate the time taken using the formula:
Time = Distance / Speed.

To find the distance between the two boats, we need to consider the relative motion between them. Let's calculate the relative velocity:
Relative velocity = Velocity of boat A - Velocity of boat B.

Velocity of boat A = 8 km/h in the direction of vector r2 - r1.
So, velocity of boat A = r2 - r1 = (2i + 5j) - (4i + 3j) = -2i + 2j.

Relative velocity = -2i + 2j - √2(-5i + 3j)
= -2i + 2j + √2(5i - 3j)
= (3√2 - 2)i + (2 + √2)j.

The distance between the two boats is constantly reducing, and at the time they are close, the distance will be almost zero. So, we can assume that the boats are close when the distance between them is less than a very small value, for example, 0.001 km.

Let's assume the time taken when the boats are close as T.
Distance between the boats at time T = (3√2 - 2)i + (2 + √2)j * T.

Using the formula for distance:
(3√2 - 2)i + (2 + √2)j * T = 0.001 km.

Solving this equation will give us the value of T.

c. Shortest distance apart:
The shortest distance between the two boats is the perpendicular distance from boat A's path to boat B's position. Let's calculate this distance.

At the time they are close, boat B will be at point (3√2 - 2)i + (2 + √2)j * T.

Now, the shortest distance between boat A's path and boat B's position is given by the projection of the vector from boat A's initial position (point P) to boat B's position on the unit vector in the direction of boat A's velocity (the relative velocity vector).

Shortest distance = Projection of (3√2 - 2)i + (2 + √2)j * T on (3√2 - 2)i + (2 + √2)j / |(3√2 - 2)i + (2 + √2)j|.

We can calculate this projection and the magnitude to find the shortest distance between the boats.

To answer this question, we can use basic vector operations and kinematics principles. Let's solve each part step by step:

a) The velocity of boat B can be found by adding the velocity of boat A (8 km/h) to the given velocity vector Vba.

Vb = 8 km/h + √2(-5i + 3j)
= 8 km/h - √2(5i - 3j)
= 8 km/h - √2(5i - 3j)
= 8 km/h - √2 * 5i + √2 * 3j
= -√2 * 5i + (√2 * 3 - 8) km/h

So, the velocity of boat B is approximately (-7.1i + 4.2j) km/h.

b) To find the time taken when the boats are close, we need to determine when the positions of the two boats are equal. We can set up the equation:

r1 + Vat * t = r2 + Vb * t

Substituting the given values:

4i + 3j + 8t(i) = 2i + 5j + (-√2 * 5t)i + (√2 * 3 - 8t)j

By equating the coefficients of the respective components, we get two equations:

4 + 8t = 2 - √2 * 5t
3 = 5 + (√2 * 3 - 8t)

Solving these equations, we can find the value of t.

c) The shortest distance between the two boats can be found by calculating the magnitude of the relative position vector between the two boats when they are closest.

Let's name this relative position vector as R:

R = r2 + Vb * t - (r1 + Vat * t)

Substituting the respective values:

R = (2i + 5j) + (-√2 * 5i + (√2 * 3 - 8)j)t - (4i + 3j + 8t(i))

Now, calculate the magnitude of vector R to get the shortest distance between the two boats:

Distance = |R|

I hope this explanation helps you understand how to solve the problem step by step.

a) |√2(-5i+3j)| = √2 √(5^2+3^2) = √2 √34 = √78

b) The distance at time t is the length of B-A
B = (-5√2 t, 3√2 t)
A = (4-2√2t, 2+2√2t)
B-A = (4-3√2t, 2+5√2t)
|B-A| = 2√(17t^2 - √2 t + 5)
Now find when that is minimum.