The acceleration f text{ ft/sec}^2 of a parti | vedclass

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

Question

(B, C) Given the acceleration $f = \frac{dv}{dt} = 6 - \sqrt{1.2t}$.At $t = 0$, $v = 0$.The velocity is maximum when the acceleration $f = 0$.Setting

Vedclass · Accepted Answer

$T = 30 	ext{ sec}$

Vedclass · Answer

(B, C) Given the acceleration $f = \frac{dv}{dt} = 6 - \sqrt{1.2t}$.At $t = 0$, $v = 0$.The velocity is maximum when the acceleration $f = 0$.Setting $f = 0$, we get $6 - \sqrt{1.2t} = 0 \Rightarrow \sqrt{1.2t} = 6 \Rightarrow 1.2t = 36 \Rightarrow t = \frac{36}{1.2} = 30 	ext{ sec}$.Thus, the time $T = 30 	ext{ sec}$.To find the maximum velocity $v$, we integrate the acceleration with respect to time from $t = 0$ to $t = 30$:$v = \int_{0}^{30} (6 - \sqrt{1.2} \cdot t^{1/2}) dt$$v = [6t - \sqrt{1.2} \cdot \frac{t^{3/2}}{3/2}]_{0}^{30}$$v = 6(30) - \frac{2}{3} \sqrt{1.2} \cdot (30)^{3/2}$$v = 180 - \frac{2}{3} \sqrt{1.2} \cdot 30 \sqrt{30} = 180 - 20 \sqrt{36} = 180 - 20(6) = 180 - 120 = 60 	ext{ ft/sec}$.Therefore, the maximum velocity is $v = 60 	ext{ ft/sec}$ at $T = 30 	ext{ sec}$.

The acceleration $f \text{ ft/sec}^2$ of a particle after a time $t \text{ sec}$ starting from rest is given by $f = 6 - \sqrt{1.2t}$. Then the maximum velocity $v$ and time $T$ to attain this velocity are:

Similar Questions

The number of values of $x$ where the function $f(x) = \cos x + \cos (\sqrt{2} x)$ attains its maximum is

The absolute maximum value of the function $f(x) = 2x^3 - 3x^2 - 36x + 9$ defined on $[-3, 3]$ is

If $p(x)$ is a polynomial of degree three that has a local maximum value $8$ at $x=1$ and a local minimum value $4$ at $x=2$,then $p(0)$ is equal to:

If $f(x)=p x^3+q x^2+r x+t$ attains local minimum and local maximum values at $x=-2$ and $x=2$ respectively and $p$ is a root of $9 x^2-1=0$,then $p+q+r=$

The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module