A particle moves in a straight line so that i | vedclass

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

Question

(C) Given the displacement equation: $x^2 = 1 + t^2$.Differentiating both sides with respect to $t$: $2x \frac{dx}{dt} = 2t$,which simplifies to $x v

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given the displacement equation: $x^2 = 1 + t^2$.Differentiating both sides with respect to $t$: $2x \frac{dx}{dt} = 2t$,which simplifies to $x v = t$,where $v$ is the velocity.Differentiating $x v = t$ with respect to $t$ using the product rule: $x \frac{dv}{dt} + v \frac{dx}{dt} = 1$.Since $\frac{dv}{dt} = a$ (acceleration) and $\frac{dx}{dt} = v$,we get $x a + v^2 = 1$.Substituting $v = \frac{t}{x}$ into the equation: $x a + (\frac{t}{x})^2 = 1$.$x a = 1 - \frac{t^2}{x^2} = \frac{x^2 - t^2}{x^2}$.Since $x^2 - t^2 = 1$ from the original equation,we have $x a = \frac{1}{x^2}$.Therefore,$a = \frac{1}{x^3} = x^{-3}$.Comparing this with $x^{-n}$,we find $n = 3$.

$A$ particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2 = 1 + t^2$. Its acceleration at any time $t$ is $x^{-n}$ where $n = . . . . .$

Similar Questions

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

The average velocity of a body moving with uniform acceleration travelling a distance of $3.06 \ m$ is $0.34 \ ms^{-1}$. If the change in velocity of the body is $0.18 \ ms^{-1}$ during this time,its uniform acceleration is ......... $ms^{-2}$.

The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,then what is the retardation?

The stopping distance of a vehicle moving with a given acceleration is proportional to the ....... of its speed.

If a body starts from rest and travels $120 \,cm$ in the $6^{th}$ second,what is the acceleration in $m/s^2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module