A point moves such that its displacement as a | vedclass

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

Question

(B) Given the displacement equation: $x^3 = t^3 + 1$. Differentiating both sides with respect to time $t$: $3x^2 \frac{dx}{dt} = 3t^2$ $x^2 v = t^2$,w

Vedclass · Accepted Answer

$\frac{2t}{x^5}$

Vedclass · Answer

(B) Given the displacement equation: $x^3 = t^3 + 1$. Differentiating both sides with respect to time $t$: $3x^2 \frac{dx}{dt} = 3t^2$ $x^2 v = t^2$,where $v = \frac{dx}{dt}$ is the velocity. $v = \frac{t^2}{x^2}$. Differentiating $x^2 v = t^2$ with respect to time $t$ using the product rule: $2x \frac{dx}{dt} v + x^2 \frac{dv}{dt} = 2t$ Since $a = \frac{dv}{dt}$,we have: $2x v^2 + x^2 a = 2t$ Substitute $v = \frac{t^2}{x^2}$: $2x (\frac{t^2}{x^2})^2 + x^2 a = 2t$ $2x \frac{t^4}{x^4} + x^2 a = 2t$ $x^2 a = 2t - \frac{2t^4}{x^3}$ Since $x^3 = t^3 + 1$,we have $t^3 = x^3 - 1$: $x^2 a = 2t - \frac{2t(t^3)}{x^3} = 2t - \frac{2t(x^3 - 1)}{x^3} = 2t - 2t + \frac{2t}{x^3} = \frac{2t}{x^3}$ $a = \frac{2t}{x^3 \cdot x^2} = \frac{2t}{x^5}$.

$A$ point moves such that its displacement as a function of time is given by $x^3 = t^3 + 1$. Its acceleration as a function of time $t$ will be

Similar Questions

The distance $x$ covered in time $t$ by a body having initial velocity $u$ and having constant acceleration $a$ is given by $x=ut+\frac{1}{2}at^2$. This result follows from

$A$ body is moving from rest under constant acceleration. Let $S_1$ be the displacement in the first $(p - 1) \ s$ and $S_2$ be the displacement in the first $p \ s$. The displacement in the $(p^2 - p + 1)^{th} \ s$ will be:

The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} \ m/s^2$ in the third second is:

$A$ train moves from rest with a uniform acceleration $a$. After attaining a maximum speed $v$,it starts moving with uniform retardation $a$. Assuming $s$ is the total distance covered in the unidirectional motion of the train,find its total time of journey and maximum speed.

Vedclass Test Series

Exam Paper Generator

Online Exam Module