A particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 -3t^2 + 2)\,m$ The displacement when the acceleration becomes zero is........$m$
$0$
$2$
$3$
$-2$
An object, moving with a speed of $6.25\, m/s$, is decelerated at a rate given by
$\frac{{dv}}{{dt}} = - 2.5\sqrt v$
where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be ........$s$
The accompanying graph of position $x$ versus time $t$ represents the motion of a particle. If $p$ and $q$ are both positive constants, the expression that best describes the acceleration $a$ of the particle is
Equation of motion of a body is $\frac{d v}{d t}=-4 v+8$, where $v$ is the velocity in $m / s$ and $t$ is the time in second. Initial velocity of the particle was zero. Then,
A body is at rest at $x=0$. At $t=0$, it starts moving in the positive $x-$ direction with a constant acceleration. At the same instant another body passes through $x=0$ moving in the positive $x$ direction with a constant speed. The position of the first body is given by $x_{1} (t)$ after time $t$ and that of the second body by $x_{2}(t)$ after the same time interval. Which of the following graphs correctly describe $\left(x_{1}-x_{2}\right)$ as a function of time $t$?
What would be the stopping distance if the velocity of vehicle becomes three times ?