5. Two particles P and Q are moving with constant velocities of (i + j

1.	5. Two particles P and Q are moving with constant velocities of (i + j) m/s and (-i + 2j) m/s respectively. At time t = 0, P is at origin and Q is at a point with position vector (2i + j)m. If the equation of the trajectory of Q as observed by P is x + 2y = n, then find n.6. A boat moves relative to water with a velocity half of the river flow velocity. If the angle from the direction of flow at which the boat must move relative to stream direction to minimize drift is 2π/n, then find n.7. A swimmer crosses the river along the line making an angle of 45 with the direction of flow. Velocity of the river water is 5m/s. Swimmer takes 6 second to cross the river of width 60m if the velocity of swimmer with respect to water is 5√n m/s. Then find n.
Answer» 5. The value of n will be 4. Explanation: As given, velocity of point P is V_p = (i + j) The velocity of Point Q is V_q = (-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. v_x = 2v - vsinϴ v_y = vcosϴ Let width of the river be d. So, Time required to cross the river,t d/vcos(ϴ). Drift ,x = v_x 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

5. Two particles P and Q are moving with constant velocities of (i + j) m/s and (-i + 2j) m/s respectively. At time t = 0, P is at origin and Q is at a point with position vector (2i + j)m. If the equation of the trajectory of Q as observed by P is x + 2y = n, then find n.6. A boat moves relative to water with a velocity half of the river flow velocity. If the angle from the direction of flow at which the boat must move relative to stream direction to minimize drift is 2π/n, then find n.7. A swimmer crosses the river along the line making an angle of 45 with the direction of flow. Velocity of the river water is 5m/s. Swimmer takes 6 second to cross the river of width 60m if the velocity of swimmer with respect to water is 5√n m/s. Then find n.

Answer»

5. The value of n will be 4.

Explanation:

As given, velocity of point P is V_p = (i + j)
The velocity of Point Q is V_q = (-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.
v_x = 2v - vsinϴ
v_y = vcosϴ
Let width of the river be d.
So, Time required to cross the river,t d/vcos(ϴ).
Drift ,x = v_x 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

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