The path of a particle moving under the influence of a force that is fixed in magnitude and direction is:

The stopping distance of a vehicle moving with a given acceleration is proportional to the ....... of its speed.

$A$ 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

$A$ lift performs the first part of its ascent with uniform acceleration $a$ and the remaining with uniform retardation $2a$. If $t$ is the total time of ascent,find the total height of the shaft.

The distance travelled by a particle is proportional to the square of time. Then,the particle travels with:

The velocity $v$ of a particle as a function of its position $x$ is expressed as $v = \sqrt{c_1 - c_2 x}$,where $c_1$ and $c_2$ are positive constants. The acceleration of the particle is