A particle is moving along positive x-axis. Its position varies as x=t3-3t2+12t+20, where x is in meters and t is in seconds.

Initial acceleration of the particle is

(A)  Zero

(B)  1 m/s2

(C)  -3 m/s2

(D)  -6 m/s2

Subtopic:  Differentiation |
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Two forces F1=2i^+2j^ N and F2=3j^+4k^ N are acting on a particle.

The resultant force acting on particle is:

(A)  2i^+5j^+4k^

(B)  2i^-5j^-4k^

(C)  i^-3j^-2k^

(D)  i^-j^-k^

Subtopic:  Resultant of Vectors |
 84%
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A=4i+4j-4k and B=3i+j+4k, then angle between vectors A and B is:

(1)  180°

(2)  90°

(3)  45°

(4)  0°

Subtopic:  Resultant of Vectors |
 77%
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If a curve is governed by the equation y=sinx, then the area enclosed by the curve and x-axis between x =0 and x =π is (shaded region) :

              

1. 1 unit

2. 2 units

3. 3 units

4. 4 units

Subtopic:  Integration |
 58%
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The acceleration of a particle starting from rest varies with time according to relation, a=α t+β. The velocity of the particle at time instant t is: (\(Here, a=\frac{dv}{dt}\))

1. αt2+βt

2. αt2+βt2

3. αt22+βt

4. 2αt2+βt

Subtopic:  Integration |
 85%
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The displacement of the particle is zero at t=0 and at t=t it is x. It starts moving in the x-direction with a velocity that varies as v=kx, where k is constant. The velocity will : (Here, \(v=\frac{dx}{dt}\))

1. vary with time.

2. be independent of time.

3. be inversely proportional to time.

4. be inversely proportional to acceleration.

Subtopic:  Integration |
 51%
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The acceleration of a particle is given as a=3x2.  At t = 0, v = 0 and x = 0. It can then be concluded that the velocity at t = 2 sec will be: (Here, \(a=v\frac{dv}{dx}\))

1.  0.05 m/s

2. 0.5 m/s

3. 5 m/s

4. 50 m/s 

Subtopic:  Integration |
 62%
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The acceleration of a particle is given by a=3t  at t=0, v=0, x=0. The velocity and displacement at t = 2 sec will be: (\(Here, a=\frac{dv}{dt}~ and~v=\frac{dx}{dt}\))

1. 6 m/s, 4 m

2. 4 m/s, 6 m

3. 3 m/s, 2 m

4. 2 m/s, 3 m

Subtopic:  Integration |
 85%
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The 9 kg block is moving to the right with a velocity of 0.6 m/s on a horizontal surface when a force F, whose time variation is shown in the graph, is applied to it at time t = 0. Calculate the velocity v of the block when t= 0.4s. The coefficient of kinetic fricton is μk=0.3[This question includes concepts from Work, Energy & Power chapter]

1. 0.6 m/s

2. 1.2 m/s

3. 1.8 m/s

4. 2.4 m/s

Subtopic:  Friction |
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The relationship between force and position is shown in the figure given (in one dimensional case). Find the work done by the force in displaying a body from x= 1 cm to x= 5cm is [This question includes concepts from Work, Energy and Power chapter]

1. 10 erg

2. 20 erg

3. 30 erg

4. 40 erg

Subtopic:  Concept of Work |
 75%
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