The position of a point at time t is given by | vedclass

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

Question

(A) Given the position coordinates: $x = a + bt - ct^2$ and $y = at + bt^2$. To find the acceleration,we first find the velocity components by differe

Vedclass · Accepted Answer

$2 \sqrt{b^2 + c^2} 	ext{ unit/s}^2$

Vedclass · Answer

(A) Given the position coordinates: $x = a + bt - ct^2$ and $y = at + bt^2$. To find the acceleration,we first find the velocity components by differentiating with respect to time $t$: $v_x = \frac{dx}{dt} = b - 2ct$ $v_y = \frac{dy}{dt} = a + 2bt$ Now,we find the acceleration components by differentiating velocity with respect to time $t$: $a_x = \frac{dv_x}{dt} = -2c$ $a_y = \frac{dv_y}{dt} = 2b$ The resultant acceleration $a$ is given by the magnitude of the acceleration vector: $a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-2c)^2 + (2b)^2}$ $a = \sqrt{4c^2 + 4b^2} = 2 \sqrt{c^2 + b^2} 	ext{ unit/s}^2$.

The position of a point at time $t$ is given by $x = a + bt - ct^2$ and $y = at + bt^2$. The resultant acceleration of the point at time $t$ is given by:

Similar Questions

If $x=4 \cos ^3 \theta$ and $y=3 \sin ^2 \theta$,then $\frac{d^2 y}{d x^2}$ at $\theta=\frac{\pi}{4}$ is

If $x=\sec \theta-\cos \theta$,$y=\sec ^{10} \theta-\cos ^{10} \theta$ and $(x^2+4)(\frac{dy}{dx})^2=k(y^2+4)$,then $k=$

For $a > 0, t \in \left( 0, \frac{\pi}{2} \right)$,let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then,$1 + \left( \frac{dy}{dx} \right)^2$ equals

If $\theta$ is the angle made by the normal drawn to the curve $x=e^{t} \cos t, y=e^{t} \sin t$ at the point $(1,0)$,with the $X$-axis,then $\theta=$

Find the rate of change of $\sqrt{x^2 + 16}$ with respect to $\frac{x}{x - 1}$ at $x = 3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module