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:
a) i+2j
b) 2i+j
c) 3i+2j
d) 3i+j
Solution:
F1 = 1j and F2 = 2i
∴Resultant vector = F1 + F2 = 2i + 1j
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
