A point moves in a straight line so that its | 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$$\frac{dx}{dt} = \frac{t}{x

Vedclass · Accepted Answer

$\frac{1}{x} - \frac{t^2}{x^3}$

Vedclass · Answer

(C) Given the displacement equation: $x^2 = 1 + t^2$.Differentiating both sides with respect to $t$:$2x \frac{dx}{dt} = 2t$$\frac{dx}{dt} = \frac{t}{x} = v$ (velocity).Now,differentiate velocity $v$ with respect to $t$ to find acceleration $a$:$a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{t}{x} \right)$Using the quotient rule $\frac{d}{dt} (u/v) = \frac{v(du/dt) - u(dv/dt)}{v^2}$:$a = \frac{x(1) - t(\frac{dx}{dt})}{x^2}$Substitute $\frac{dx}{dt} = \frac{t}{x}$ into the equation:$a = \frac{x - t(\frac{t}{x})}{x^2}$$a = \frac{x - \frac{t^2}{x}}{x^2}$$a = \frac{x}{x^2} - \frac{t^2}{x^3}$$a = \frac{1}{x} - \frac{t^2}{x^3}$.

$A$ point moves in a straight line so that its displacement $x \ m$ at time $t \ s$ is given by $x^2 = 1 + t^2$. Its acceleration in $m/s^2$ at a time $t \ s$ is

Similar Questions

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:

Draw the $x-t$ graphs for positive,negative,and zero acceleration.

The displacement is given by $x = 2t^2 + t + 5$,the acceleration at $t = 2 \ s$ is ......... $m/s^2$.

The acceleration-time graph for a particle is given in the figure. If it starts motion at $t=0$, the distance travelled in $3 \, s$ will be ........... $m$.

$A$ particle starts from rest. Its acceleration $(a)$ versus time $(t)$ graph is as shown in the figure. The maximum speed of the particle will be (in $m \ s^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module