## How do you solve Numericals in motion in a plane?

(a) 10i + 5j

(b) 4i - 6j

(c) 9i + 6j

(d) None of these

**Solution:**
Given, Point A = (-4, 6) and Point B = (5, 12)

Let the origin be O = (0, 0)

Position vector of A = OA = -4i + 6j

Position vector of B = OB = 5i + 12j

∴ Displacement vector = AB = OB - OA = 5i + 12j + 4i - 6j = 9i + 6j

(a) moving with uniform velocity

(b) stationary

(c) moving with uniform acceleration

(d) None of these

**Solution: **
=> Postion vector given r=2ti -4j+k

=> diffrentiating with respect to time

=> dr/dt=2i

=> velocity=2i

Particle is moving with uniform velocity along x-axis.

(a) 400

(b) 500

(c) 600

(d) 50

**Solution: **
r = 3t²i + 4t²j + 7k

Putting t = 0, we get

=> Initial position = r1 = 7k

Putting t = 10, we get

=> Final position = r2 = 300i + 400j + 7k

∴ Displacement of the particle = r2 - r1 = 300i + 400j + 7k - 7k = 300i + 400j

Magnitude of the displacement of the particle = √[(300)² + (400)²] = 500

(a) moving with uniform velocity

(b) stationary

(c) moving with uniform acceleration

(d) None of these

**Solution: **
=> Postion vector given r = 2t²i - 4j + k

=> diffrentiating with respect to time

=> dr/dt = 4t

=> velocity = v = 4ti

=> dv/dt = 4i

Particle is moving with uniform acceleration along x-axis.

(a) -3i - 2j

(b) -2i - 3j

(c) 9i + 6j

(d) None of these

**Solution:**

First he moves from O to A in North direction for 2 m.

∴Displacement vector of OA = 2j

Then he moves from A to B in East direction for 1 m.

∴Displacement vector of AB = 1i

Then he moves from B to C in South direction for 5 m.

∴Displacement vector of BC = -5j

Then he moves from C to D in West direction for 3 m.

∴Displacement vector of CD = -3i

So, the displacment vector of OD = 2j + 1i - 5j - 3i = -2i - 3j

(a) -3i - 2j

(b) -2i - 3j

(c) (√(3) + 2)i - 2j

(d) None of these

**Solution:**

First he moves from A to B in North - West direction at 30° for 2 m.

∴Displacement vector of AB = 2cos30°i + 2sin30°j = √(3)i + j

Then he moves from B to C in East direction for 2 m.

∴Displacement vector of BC = 2i

Then he moves from C to D in South direction for 3 m.

∴Displacement vector of CD = -3j

So, the displacment vector of AD = √(3)i + j + 2i - 3j = (√(3) + 2)i - 2j

(a) 10 m along North

(b) 10 m along South

(c) 10 m along West

(d) None of these

**Solution:**

First he moves from O to A in North direction for 30 m.

∴Displacement vector of OA = 30j

Then he moves from A to B in East direction for 20 m.

∴Displacement vector of AB = 20i

Then he moves from B to C in South - West direction for 30√2 m.

∴Displacement vector of BC = (30√2cos45°)i + (30√2sin45°)j = -30i - 30j

So, the displacment vector of OC = 30j + 20i - 30i - 30j = -10i

(a) Find vector expression for the particle position as a function of time, using the unit vectors i and j

(b) Determine the expression for the velocity vector as a function of time

(c) Obtain the expression for the acceleration vector a as a function of time.

(d) Find the the position, the velocity, and the acceleration of the particle at t = 1s

b) velocity = 22 - 9.8t

c) acceleration = - 9.8 ( It is not the function of t. It is constant retardation)

d) Position vector at 1 sec = 18i - 0.9j, Velocity at 1 sec = 12.2 unit/sec

**Solution:**
Given, x = (18)t and y = 4t - (4.9)t²

a) Position vector at t time (r)= 18ti + [4t - (4.9)t²]j = 18ti + 4tj - (4.9)t²j

b) velocity (v)= dr/dt= 22 - 9.8t

c) acceleration (a)= dv/dt= - 9.8 ( It is not the function of t. It is constant retardation)

d) Position vector at 1 sec = 18i - 0.9j, Velocity at 1 sec = 12.2 unit/sec

a) 16 m/s²

b) 12 m/s²

c) 54 m/s²

d) 20 m/s²

**Solution:**
Given, x(t)=4t³ +3t²+2

velocity (v)=dx/dt=12t²+6t

Acceleration (a)=dv/dt= 24t+6

therefore At t=2 sec ,

Acceleration (a)= 24x2+6=54

i) Initial velocity of the particle is u.

ii) Acceleration of the particle is a.

iii) Acceleration of the particle is 2a.

iv) At t-2 sec, The particle is at orign.

a) i

b) ii

c) iii & iv

d) None of these

**Solution:**
Given, x=u(t−2)+a(t−2)²

At t=2 sec ,

x=u(2−2)+a(2−2)²

x=0

At t-2 sec, The particle is at orign.

Given, x=u(t−2)+a(t−2)²

x=ut-2u+a(t²+4-4t)

x=ut-2u+at²+4a-4at)

dx/dt=v=2at-4a

dv/dt=2a

Acceleration of the particle is 2a.

So, (iii & iv) statements are true.

a) 1

b) 2

c) 4

d) None of these

**Solution:**
Given, v=10+5t-t²

Acceleration (a)=dv/dt= 5-2t

therefore At t=2 sec ,

Acceleration (a)= 5-4=1

a)

b)

c)

d) None of these

**Solution:**

F1=-4i-5j-5k,

F2= 5i+8j+6k,

F3=-3i+4j-7k,

F4= 2i-3j-2k

Then, particle will move in.........Plane.

a) x-y

b) y-z

c) x-z

d) Along x axis

**Solution:**
Given, F1=-4i-5j-5k, F2= 5i+8j+6k, F3=-3i+4j-7k, F4= 2i-3j-2k

Resultant force = r = F1 + F2 + F3 + F4 = (-4i-5j-5k) + (5i+8j+6k) + (-3i+4j-7k) + (2i-3j-2k) = 0i + 4j - 8k

So, the object will move in y-z plane.

a) 30°,60°,90°

b) 45°,45°,90°

c) 45°,60°,90°

d) 90°,135°,135°

**Solution:**
Given, A + B + C = 0

Let, one side of the triangle be x.

So the second side will also be x.

And the third side is x√2.

This is only possible when the triangle is right angled isosecles trinagle.

a) 30°

b) 45°

c) 90°

d) 120°

**Solution:**
Let two unit vectors a, and b, whose sum is also a unit vector c and angle between them is θ.

a + b = c and |a| = |b| = |c| = 1

Now, we know that magnitude of the sum of vector a and vector b is:

|a + b|² = |a|² + |b|² + 2 |a||b|cosθ

Substituting the values we get:

1 = 1 + 1 + 2 (1x1 cosθ)

cosθ = -½ = cos 120°

θ = 120°

a) 0

b) √3

c) 2

d) None of these

**Solution:**
Let two unit vectors a, and b, whose sum is also a unit vector c and angle between them is θ

a + b = c and |a| = |b| = |c| = 1

Now, we know that magnitude of the sum of vector a and vector b is:

|a + b|² = |a|² + |b|² + 2 |a||b|cosθ

Substituting the values we get:

1 = 1 + 1 + 2 (|a||b|cosθ)

|a||b|cosθ = -½ ---------------Eqn 1

Magnitude of their difference is given by:

|a - b|² = |a|² + |b|² - 2 |a||b|cosθ

Substituting the values |a||b|cosθ from Eqn 1 we get:

|a - b|²=1 + 1 - 2 x -½ = 3

|a - b| = √3

(a) 8i-6j+3k

(b) 3i-6j+8k

(c)-3i-6j+6k

(d) 8i-6j

**Solution: **
=> v=ωxr

=> v= (3i+4j)x(j+2k)

=> v= 8i-6j+3k