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 $\text{ m/s}^2$)

When will the average acceleration and instantaneous acceleration of a particle be equal over any time interval?

For a particle moving along a straight line path, the displacements in the third and fifth seconds of its motion are $10 \ m$ and $18 \ m$ respectively. The speed of the particle at time $t=4 \ s$ is (in $ms^{-1}$)

The displacement $(x)$ of a particle depends on time $t$ as $x = \alpha t^2 - \beta t^3$. Choose the incorrect statements from the following.

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

$A$ car starts at time $t=0$ from an initial speed of $10 \,m/s$ and accelerates uniformly with $2 \,m/s^2$ on a straight road for time $0 \leq t \leq 10 \,s$. Let $S_1$ and $S_2$ be the distance covered by the car in time $3 \leq t \leq 4 \,s$ and $4 \leq t \leq 5 \,s$ respectively. The ratio $\frac{S_2}{S_1}$ is