A particle moving in a straight line starts f | vedclass

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

Question

(A) Given acceleration $v'(t) = a - kt^2$. Since the particle starts from rest,$v(0) = 0$.Integrating with respect to $t$,we get $v(t) = \int (a - kt^

Vedclass · Accepted Answer

$\frac{2}{3} \sqrt{\frac{a^3}{k}}$

Vedclass · Answer

(A) Given acceleration $v'(t) = a - kt^2$. Since the particle starts from rest,$v(0) = 0$.Integrating with respect to $t$,we get $v(t) = \int (a - kt^2) dt = at - \frac{k}{3}t^3 + C$.Using $v(0) = 0$,we find $C = 0$,so $v(t) = at - \frac{k}{3}t^3$.To find the maximum velocity,we set the acceleration to zero: $a - kt^2 = 0$,which gives $t^2 = \frac{a}{k}$,or $t = \sqrt{\frac{a}{k}}$ (since $t > 0$).Substituting this value of $t$ into the velocity equation:$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \left(\sqrt{\frac{a}{k}}\right)^3$$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \cdot \frac{a}{k} \sqrt{\frac{a}{k}}$$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{a}{3} \sqrt{\frac{a}{k}} = \frac{2a}{3} \sqrt{\frac{a}{k}} = \frac{2}{3} \sqrt{\frac{a^3}{k}}$.

$A$ particle moving in a straight line starts from rest and the acceleration at any time $t$ is $a - kt^2$,where $a$ and $k$ are positive constants. The maximum velocity attained by the particle is

Similar Questions

If the function $f(x)=2x^3-9ax^2+12a^2x+1, a>0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$,then $\alpha$ and $\alpha^2$ are the roots of the equation:

Let $f: R \to R$ be defined by $f(x) = \begin{cases} k - 2x, & \text{if } x \leqslant -1 \\ 2x + 3, & \text{if } x > -1 \end{cases}$. If $f$ has a local minimum at $x = -1$,then what is the possible value of $k$?

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius $3$ is

Let $x=2$ be a local minima of the function $f(x)=2x^4-18x^2+8x+12$,$x \in (-4,4)$. If $M$ is the local maximum value of the function $f$ in $(-4,4)$,then $M =$

The minimum value of the expression $7 - 20x + 11x^2$ is (in $/11$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module