A street car moves rectilinearly from station | vedclass

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

Question

(A) Given acceleration $a = b - cx$.We know that $a = v \frac{dv}{dx}$.So,$v \frac{dv}{dx} = b - cx$.Integrating both sides: $\int v \, dv = \int (b -

Vedclass · Accepted Answer

$x = 2b/c, v_{\max} = b/\sqrt{c}$

Vedclass · Answer

(A) Given acceleration $a = b - cx$.We know that $a = v \frac{dv}{dx}$.So,$v \frac{dv}{dx} = b - cx$.Integrating both sides: $\int v \, dv = \int (b - cx) \, dx$.$\frac{v^2}{2} = bx - \frac{cx^2}{2} + C$. Since at $x = 0$,$v = 0$,the constant $C = 0$.Thus,$v^2 = 2bx - cx^2$.At the next station $B$,the velocity $v = 0$.$0 = x(2b - cx) \implies x = 2b/c$ (since $x=0$ is station $A$).For maximum velocity,$\frac{dv}{dx} = 0$. Differentiating $v^2 = 2bx - cx^2$ with respect to $x$: $2v \frac{dv}{dx} = 2b - 2cx$.Setting $\frac{dv}{dx} = 0$,we get $2b - 2cx = 0$,so $x = b/c$.Substituting $x = b/c$ into the expression for $v^2$:$v_{\max}^2 = 2b(b/c) - c(b/c)^2 = 2b^2/c - b^2/c = b^2/c$.Therefore,$v_{\max} = b/\sqrt{c}$.

$A$ street car moves rectilinearly from station $A$ to the next station $B$ with an acceleration varying according to the law $a = (b - cx)$,where $b$ and $c$ are constants and $x$ is the distance from station $A$. The distance between the two stations and the maximum velocity are:

Similar Questions

In what different ways can the velocity of an object be changed?

Which one of the following statements is wrong?

$A$ particle moves along the sides $AB, BC, CD$ of a square of side $25 \, m$ with a constant velocity of $15 \, m \, s^{-1}$. Its average velocity is ........ $m \, s^{-1}$.

$A$ ball moves one-fourth of a circle of radius $R$ in time $T$. Let $v_1$ and $v_2$ be the magnitudes of mean speed and mean velocity vector. The ratio $\frac{v_1}{v_2}$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module