The distance travelled by a particle moving i | vedclass

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

Question

(C) Given the distance $s = \sqrt{at^2 + bt + c}$,we have $s^2 = at^2 + bt + c$. Differentiating with respect to $t$: $2s \frac{ds}{dt} = 2at + b$ $\f

Vedclass · Accepted Answer

Proportional to $s^{-3}$

Vedclass · Answer

(C) Given the distance $s = \sqrt{at^2 + bt + c}$,we have $s^2 = at^2 + bt + c$. Differentiating with respect to $t$: $2s \frac{ds}{dt} = 2at + b$ $\frac{ds}{dt} = \frac{2at + b}{2s}$ Differentiating again with respect to $t$ to find acceleration $f = \frac{d^2s}{dt^2}$: $2 \left( \frac{ds}{dt} \right)^2 + 2s \frac{d^2s}{dt^2} = 2a$ $s \frac{d^2s}{dt^2} = a - \left( \frac{ds}{dt} \right)^2$ $s \frac{d^2s}{dt^2} = a - \frac{(2at + b)^2}{4s^2} = \frac{4as^2 - (2at + b)^2}{4s^2}$ Substituting $s^2 = at^2 + bt + c$: $s \frac{d^2s}{dt^2} = \frac{4a(at^2 + bt + c) - (4a^2t^2 + 4abt + b^2)}{4s^2}$ $s \frac{d^2s}{dt^2} = \frac{4a^2t^2 + 4abt + 4ac - 4a^2t^2 - 4abt - b^2}{4s^2} = \frac{4ac - b^2}{4s^2}$ $\frac{d^2s}{dt^2} = \frac{4ac - b^2}{4s^3}$ Since $4ac - b^2$ is a constant,the acceleration $f \propto s^{-3}$.

The distance travelled by a particle moving in a straight line in time $t$ is $s = \sqrt{at^2 + bt + c}$. The acceleration of the particle is

Similar Questions

From a balloon rising vertically with a uniform velocity of $v \ ft/sec$,a stone is dropped. If the stone reaches the ground after $4 \ sec$,what is the height of the balloon above the ground at that moment (in $ft$)? (Take $g = 32 \ ft/sec^2$)

$A$ particle is moving on a straight line. The distance $S$ travelled in time $t$ is given by $S = at^2 + bt + 6$. If the particle comes to rest after $4 \text{ s}$ at a distance of $16 \text{ m}$ from the starting point,then the acceleration of the particle is:

The two equal sides of an isosceles triangle with a fixed base $b$ are decreasing at the rate of $3 \ cm/s$. How fast is the area decreasing when the two equal sides are equal to the base?

The rate of change of the volume of a sphere with respect to its surface area,when the radius is $5 \ m$ is

The radius of a right circular cylinder increases at the rate of $0.1 \text{ cm/min}$,and the height decreases at the rate of $0.2 \text{ cm/min}$. The rate of change of the volume of the cylinder,in $\text{cm}^3/\text{min}$,when the radius is $2 \text{ cm}$ and the height is $3 \text{ cm}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module