5. The value of n will be 4.
Explanation:
As given, velocity of point P is Vp = (i + j)
The velocity of Point Q is Vq = (-i + 2j)
Displacement of point P is x1i + y1j = (i + j) × t
Displacement of point Q is (x2 − 2)i + (y2 - 1)j = (−2i + j) × t
Displacement of point Q with respect to P is
[(x2 − 2)i + (y2 - 1)j] - (x1i + y1j) = (-i + 2j) x t - (i + j) x t
(x2 − x1)i + (y2 - y1)j - 2i - j = (-i + 2j) x t - (i + j) x t
(x − 2)i + (y - 1)j = (−2 it + tj) [as (x2 − x1) = x and (y2 − y1) = y]
x − 2 = −2t _______(i)
and y − 1 = t _______(ii)
x − 2 = −2(y − 1)
x − 2 = −2y + 2
2y + x = 4
comparing with the given eqn n = 4
6. The value of n will be 3.
Explanation:
Let the relative velocity of boat be 'v' m/s making an angle ϴ with vertical.
Then,the velocity of river is '2v' in hrizontal direction.
Now, break velocity of boat in respective components.
vx = 2v - vsinϴ
vy = vcosϴ
Let width of the river be d.
So, Time required to cross the river,t d/vcos(ϴ).
Drift ,x = vx x t = 2v - vsinϴ x d/vcos(ϴ)
For min. drift,
dx/dϴ = 0
d/dϴ [d/v(2vsecϴ - v tanϴ)]
2vtanϴ - v secϴ = 0
sin ϴ = 1/2
ϴ = 30
This angle is the angle made with the vertical. Therefore the total angle made with the direction of river flow is 90+30=120 degree which is equal to 2π/3.
Hence, n = 3.
7. The value of n is 5.
Explanation:
Let the velocity of the swimmer be v m/s.
So, velocity along vertical direction is v sin45°
Total width = 60 m and time taken = 6 s
So, v sin45° = 60/6
v/√2 = 10
v = 10√2
or in the other way v = 10i + 100
thus, velocity of swimmer w.r.t river = 10i + 10j - 5i
= |5i + 10j|
=√25 + 100
hence n = 5