The acceleration \(a\) (in ${\mathrm{ms}}^{-2}$) of a body, starting from rest varies with time \(t\) (in \(\mathrm{s}\)) as per the equation \(a=3t+4.\) The velocity of the body at time \(t=2\) \(\mathrm{s}\) will be:
1. \(10\) ${\mathrm{ms}}^{-1}$
2. \(18\) ${\mathrm{ms}}^{-1}$
3. \(14\) ${\mathrm{ms}}^{-1}$
4. \(26\) ${\mathrm{ms}}^{-1}$

A body thrown vertically so as to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is:

1. $\sqrt{2}t$

2. $\left(1+\frac{1}{\sqrt{2}}\right)t$

3. $\frac{3t}{2}$

4. $\frac{t}{\sqrt{2}}$

A stone falls freely from rest from a height h and it travels a distance $\frac{9h}{25}$ in the last second. The value of h is:

1. 145 m

2. 100 m

3. 125 m

4. 200 ms

A point moves in a straight line under the retardation a${\mathrm{v}}^2$. If the initial velocity is \(\mathrm{u},\) the distance covered in \(\mathrm{t}\) seconds is:
1. $\mathrm{aut}$
2. $\frac{1}{\mathrm{a}}\ln$ $\left(\mathrm{aut}\right)$
3. $\frac{1}{\mathrm{a}}\ln$ $\left(1+\mathrm{aut}\right)$
4. $\mathrm{a}$ $\ln$ $\left(\mathrm{aut}\right)$

A particle is thrown upwards from ground. It experiences a constant resistance force which can produce retardation of 2 $m/s^2$. The ratio of time of ascent to the time of descent is: $[g=10 m/s^2]$

(1) 1:1

(2) $\sqrt{\frac{2}{3}}$

(3) $\frac{2}{3}$

(4) $\sqrt{\frac{3}{2}}$

A bullet loses $\frac{1}{20}$ of its velocity passing through a plank. The least number of planks required to stop the bullet is (All planks offers same retardation)

(1) 10

(2) 11

(3) 12

(4) 23

A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by $(4t^3-2t)$, where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin ?

1. 28 m/s²

2. 22 m/s²

3. 12 m/s²

4. 10 m/s²

The relation between time and distance is given by $t=\alpha x^2+\beta x$, where α and β are constants. The retardation, as calculated based on this equation, will be (assume v to be velocity) :

1. $2\alpha v^3$

2. $2\beta v^3$

3. $2\alpha \beta v^3$

4. $2{\beta }^2{v}^3$

A point moves with uniform acceleration and v₁, v₂ and v₃ denote the average velocities in the three successive intervals of time t₁, t₂ and t₃. Which of the following relations is correct ?

(1) $(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2+t_3)$

(2) $(v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$

(3) $(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_1-t_3)$

(4) $(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2-t_3)$ 

The acceleration of a moving body can be found from: 

(1) Area under the velocity-time graph

(2) Area under the distance-time graph

(3) Slope of the velocity-time graph

(4) Slope of the distance-time graph