A particle moves in a straight line and its p | vedclass

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

Question

(B) Given the position equation: $x^2 = t + 2$.Differentiating both sides with respect to time $t$:$2x \frac{dx}{dt} = 1 \Rightarrow \frac{dx}{dt} = \

Vedclass · Accepted Answer

$-\frac{1}{4x^3}$

Vedclass · Answer

(B) Given the position equation: $x^2 = t + 2$.Differentiating both sides with respect to time $t$:$2x \frac{dx}{dt} = 1 \Rightarrow \frac{dx}{dt} = \frac{1}{2x}$.Now,differentiate the velocity $v = \frac{dx}{dt} = \frac{1}{2} x^{-1}$ with respect to time $t$ to find acceleration $a$:$a = \frac{dv}{dt} = \frac{d}{dt} (\frac{1}{2} x^{-1}) = \frac{1}{2} (-1) x^{-2} \frac{dx}{dt}$.Substitute $\frac{dx}{dt} = \frac{1}{2x}$ into the equation:$a = -\frac{1}{2x^2} 	imes \frac{1}{2x} = -\frac{1}{4x^3}$.

$A$ particle moves in a straight line and its position $x$ at time $t$ is given by $x^2 = 2 + t$. Its acceleration is given by

Similar Questions

If the velocity of a particle moving along the $x-$ axis is given by $v = k\sqrt{x}$,then which of the following is true? ($a$ is acceleration)

For a moving body at any instant of time,which of the following statements is correct?

$A$ lift performs the first part of its ascent with uniform acceleration $a$ and the remaining with uniform retardation $2a$. If $t$ is the total time of ascent,find the total height of the shaft.

The displacement of a body in $8\,s$ starting from rest with an acceleration of $20\,cm/s^2$ is:

$A$ bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further will it penetrate before coming to rest,assuming that it faces constant resistance to motion? (in $cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module