The velocity of a particle moving in the posi | vedclass

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

Question

(A) Given the velocity $v = 5 \sqrt{x}$.We know that acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.First,find

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(A) Given the velocity $v = 5 \sqrt{x}$.We know that acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v = 5x^{1/2}$ with respect to $x$:$\frac{dv}{dx} = 5 \cdot \frac{1}{2} x^{-1/2} = \frac{2.5}{\sqrt{x}}$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = (5 \sqrt{x}) \cdot \left( \frac{2.5}{\sqrt{x}} \right) = 5 \cdot 2.5 = 12.5 \, m/s^2$.Thus,the acceleration of the particle is $12.5 \, m/s^2$.

The velocity of a particle moving in the positive direction of the $x$-axis varies as $v = 5 \sqrt{x}$. Assuming that at $t = 0$,the particle was at $x = 0$,what is the acceleration of the particle in $m/s^2$?

Similar Questions

$A$ bullet fired into a target loses one-third of its velocity after travelling a distance $x$ meters into the target. If the bullet comes to rest by travelling a further distance $x^{\prime}$, then the ratio $\frac{x^{\prime}}{x}$ is

For a particle moving along a straight line path, the displacements in the third and fifth seconds of its motion are $10 \ m$ and $18 \ m$ respectively. The speed of the particle at time $t=4 \ s$ is (in $ms^{-1}$)

$A$ body starts from rest with uniform acceleration. If its velocity after $n^{\text{th}}$ second (last second) is $V$, then its displacement in the last two seconds is

The acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.

What is stopping distance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module