If the velocity of a particle moving along a straight line with uniform acceleration is given by $V = \sqrt{196 - 16x} \text{ m/s}$, then its acceleration is ($x$ is the displacement of the particle). (in $\text{ m/s}^2$)

$A$ driver applies the brakes on seeing the red traffic signal $400 \ m$ ahead. At the time of applying brakes,the vehicle was moving with $15 \ m/s$ and retarding at $0.3 \ m/s^2$. The distance of the vehicle from the traffic light one minute after the application of brakes is: (in $m$)

$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

$A$ body moving with uniform acceleration travels a distance of $25 \ m$ in the fourth second and $37 \ m$ in the sixth second. The distance covered by the body in the next two seconds is: (in $m$)

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

The relation between position $(x)$ and time $(t)$ is given below for a particle moving along a straight line. Which of the following equations represents uniformly accelerated motion? [where $\alpha$ and $\beta$ are positive constants]