The relation between the displacement x (in m | vedclass

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

Question

(B) Given the relation: $t = 2x^2 + 3x$. To find velocity $v$, differentiate $t$ with respect to $x$: $\frac{dt}{dx} = 4x + 3$. Since $v = \frac{dx}{d

Vedclass · Accepted Answer

$-\frac{1}{16} \ ms^{-2}$

Vedclass · Answer

(B) Given the relation: $t = 2x^2 + 3x$. To find velocity $v$, differentiate $t$ with respect to $x$: $\frac{dt}{dx} = 4x + 3$. Since $v = \frac{dx}{dt}$, we have $v = \frac{1}{4x + 3} = (4x + 3)^{-1}$. To find acceleration $a$, differentiate $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$. $\frac{dv}{dx} = -1(4x + 3)^{-2} \cdot 4 = -\frac{4}{(4x + 3)^2}$. Thus, $a = -\frac{4}{(4x + 3)^2} \cdot \frac{1}{4x + 3} = -\frac{4}{(4x + 3)^3}$. Given displacement $x = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$. Substitute $x = \frac{1}{4}$ into the acceleration formula: $a = -\frac{4}{(4(1/4) + 3)^3} = -\frac{4}{(1 + 3)^3} = -\frac{4}{4^3} = -\frac{4}{64} = -\frac{1}{16} \ ms^{-2}$.

The relation between the displacement $x$ (in metre) and the time $t$ (in second) of a particle is $t = 2x^2 + 3x$. If the displacement of the particle is $25 \ cm$ from the origin $(x = 0)$, then the acceleration of the particle is:

Similar Questions

Define acceleration,average acceleration,and instantaneous acceleration.

The correct position $(x)$ - time $(t)$ graph for a particle moving with negative acceleration is

The position $x$ of a particle varies with time $t$ as $x = at^2 - bt^3$. The acceleration of the particle will be zero at time $t$ equal to

$A$ particle starts from rest. Its acceleration $a$ versus time $t$ is shown in the figure. The maximum speed of the particle will be (in $m/s$)

If the displacement ($s$ in metre) of a moving particle in terms of time ($t$ in second) is $s = t^3 - 6t^2 + 18t + 9$,then the minimum velocity attained by the particle is (in $m \ s^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module