The velocity of a particle is given by the equation $v(x) = 3x^2 - 4x$,where $x$ is the distance covered by the particle. The expression for its acceleration is

$A$ particle is moving with constant acceleration,whose direction is the same as that of its instantaneous motion. The distance $(s)-$velocity $(v)$ graph for the particle is:

$A$ particle is moving in one dimension (along $x$-axis) under the action of a variable force. Its initial position was $16 \,m$ right of the origin. The variation of its position $(x)$ with time $(t)$ is given as $x = -3t^3 + 18t^2 + 16t$,where $x$ is in $m$ and $t$ is in $s$. The velocity of the particle when its acceleration becomes zero is . . . . . . $m/s$.

The velocity of a body depends on time according to the equation $v = \frac{t^2}{10} + 20$. The body is undergoing

Will the acceleration increase or decrease with an increase in velocity?

The velocity of a particle is given by $v = 2t^2 - 8t + 15 \,ms^{-1}$. Find its instantaneous acceleration at $t = 5 \,s$. (in $\,ms^{-2}$)