Consider a particle moving along the positive | vedclass

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

Question

(A) Given, $v = \alpha \sqrt{x}$.Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \alpha \sqrt{x}$.Separating variables: $\frac{dx}{\sqrt{x}} = \al

Vedclass · Accepted Answer

$\frac{\alpha^2}{2} t$ and $\frac{\alpha^2}{2}$

Vedclass · Answer

(A) Given, $v = \alpha \sqrt{x}$.Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \alpha \sqrt{x}$.Separating variables: $\frac{dx}{\sqrt{x}} = \alpha dt$.Integrating both sides: $\int x^{-1/2} dx = \int \alpha dt \Rightarrow 2\sqrt{x} = \alpha t + C$.At $t = 0$, $x = 0$, so $C = 0$. Thus, $2\sqrt{x} = \alpha t \Rightarrow \sqrt{x} = \frac{\alpha t}{2}$.Squaring both sides: $x = \frac{\alpha^2 t^2}{4}$.Velocity $v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t^2}{4} \right) = \frac{\alpha^2}{4} (2t) = \frac{\alpha^2 t}{2}$.Acceleration $a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t}{2} \right) = \frac{\alpha^2}{2}$.

Similar Questions

The velocity of a car accelerating uniformly increases from $20 \ m/s$ to $80 \ m/s$ while covering a distance of $200 \ m$. The time taken is $.... \ s$.

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

If the engine of a long train moving with constant acceleration crosses a tree with velocity $u$ and the last compartment of the train crosses the same tree with velocity $v$,then the velocity with which the middle compartment crosses the same tree is

$A$ particle is moving in a straight line with initial velocity $u$ and uniform acceleration $a$. If the sum of the distance travelled in the $t^{\text{th}}$ and $(t+1)^{\text{th}}$ seconds is $100 \text{ cm}$,then its velocity after $t$ seconds,in $\text{cm/s}$,is:

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 6t^2 + 3t + 4) \ m$. Find the velocity of the particle in $m/s$ when its acceleration is zero.

Vedclass Test Series

Exam Paper Generator

Online Exam Module