If the velocity of a particle moving along th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the velocity $v = k\sqrt{x}$.We know that acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v$ with respect

Vedclass · Accepted Answer

$a =$ constant

Vedclass · Answer

(D) Given the velocity $v = k\sqrt{x}$.We know that acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(k x^{1/2}) = k \cdot \frac{1}{2} x^{-1/2} = \frac{k}{2\sqrt{x}}$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (k\sqrt{x}) \cdot \left(\frac{k}{2\sqrt{x}}\right)$.Simplifying this,we get $a = \frac{k^2}{2}$.Since $k$ is a constant,$a$ is also a constant.

If the velocity of a particle moving along the $x-$ axis is given by $v = k\sqrt{x}$,then which of the following is true? ($a$ is acceleration)

Similar Questions

$A$ bullet is fired into a fixed target and loses one-third of its velocity after traveling $4 \ cm$. It penetrates a further $D \times 10^{-3} \ m$ before coming to rest. The value of $D$ is:

Derive the equations of motion for constant acceleration using the method of calculus.

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

The speeds of two identical cars are $u$ and $4u$ at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

The motion of a particle is described by the equation $v = at$. The distance travelled by the particle in the first $4 \ s$ is: (in $a$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module