A body is moving according to the equation $x=at+b{t}^2-c{t}^3$ where x = displacement and a, b and c are constants. The acceleration of the body is 

1. $a+2bt$ 

2. $2b+6ct$ 

3. $2b-6ct$ 

4. $3b-6c{t}^2$ 

Equation of displacement for any particle is $s=3t^3+7t^2+14t+8m$. Its acceleration at time t = 1 sec is 

1. 10 m/s²

2. 16 m/s²

3. 25 m/s²

4. 32 m/s²  

Velocity of a particle changes when 

1. Direction of velocity changes

2. Magnitude of velocity changes

3. Both of above

4. None of the above

The distance travelled by a particle is proportional to the squares of time, then the particle travels with  

1. Uniform acceleration

2. Uniform velocity

3. Increasing acceleration

4. Decreasing velocity

The relation \(3t = \sqrt{3x} + 6\) describes the displacement of a particle in one direction where \(x\) is in metres and \(t\) in seconds. The displacement, when velocity is zero, is: 

A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S₁, S₂ and S₃ in the first five seconds, second five seconds and next five seconds respectively the relation between S₁, S₂ and S₃ is 

1. $S_1=S_2=S_3$

2. $5S_1=3S_2=S_3$

3. $S_1=\frac{1}{3}{S}_2=\frac{1}{5}{S}_3$  

4. $S_1=\frac{1}{5}{S}_2=\frac{1}{3}{S}_3$

The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms^–1. If the change in velocity of the body is 0.18ms^–1 during this time, its uniform acceleration is 

1. 0.01 ms^–2

2. 0.02 ms^–2

3. 0.03 ms^–2

4. 0.04 ms^–2

The motion of a particle is described by the equation u = at, where u is velocity and a is a constant. The distance travelled by the particle in the first 4 seconds 

1. 4 a

2. 12 a

3. 6 a

4. 8 a

A particle travels 10 m in first 5 sec and 10m in the next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec ?

1. 8.3 m

2. 9.3 m

3. 10.3 m

4. None of above