A particle of unit mass undergoes one dimensional motion such that its velocity varies according to $ v(x)= \beta {x^{ - 2n}}$, where $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$, is given by
$-2n$${\beta ^2}{X^{ - 2n - 1}}$
$-2n$${\beta ^2}{X^{ - 4n - 1}}$
$-2n$${\beta ^2}{X^{ - 2n + 1}}$
$-2n$${\beta ^2}{X^{ - 4n + 1}}$
The initial velocity of a particle moving along $x$-axis is $u$ (at $t=0$ and $x=0$ ) and its acceleration $a$ is given by $a=k x$. Which of the following equation is correct between its velocity $(v)$ and position $(x)$ ?
Draw $x \to t$ graph for negative acceleration.
Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity, or deceleration, $-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0 $ and $a$.
An object with a mass $10 \,kg$ moves at a constant velocity of $10 \,m/sec$. A constant force then acts for $4\, second$ on the object and gives it a speed of $2\, m/sec$ in opposite direction. The acceleration produced in it, is ........ $m/{\sec ^2}$