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

$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

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

The velocity of a body moving with a uniform acceleration of $2\,m/s^2$ is $10\,m/s$. Its velocity after an interval of $4\,s$ is...........$m/s$.

For a moving body at any instant of time,which of the following statements is correct?

$A$ particle initially at rest is moving along a straight line with an acceleration of $2 \,m/s^2$. At a time of $3 \,s$ after the beginning of motion, the direction of acceleration is reversed. The time from the beginning of the motion in which the particle returns to its initial position is

Vedclass Test Series

Exam Paper Generator

Online Exam Module