A body starts from rest with uniform accelera | vedclass

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

Question

(C) Given that the body starts from rest,the initial velocity $u = 0$. Let the uniform acceleration be $a$.At time $t = n$,the velocity $v = u + at =

Vedclass · Accepted Answer

$\frac{2v(n-1)}{n}$

Vedclass · Answer

(C) Given that the body starts from rest,the initial velocity $u = 0$. Let the uniform acceleration be $a$.At time $t = n$,the velocity $v = u + at = 0 + an$,so $a = \frac{v}{n}$.The displacement in the $k^{	ext{th}}$ second is given by $S_k = u + \frac{a}{2}(2k - 1)$.Since $u = 0$,$S_k = \frac{a}{2}(2k - 1)$.The displacement in the $n^{	ext{th}}$ second is $S_n = \frac{a}{2}(2n - 1)$.The displacement in the $(n-1)^{	ext{th}}$ second is $S_{n-1} = \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3)$.The total displacement in these two seconds is $S_{total} = S_n + S_{n-1} = \frac{a}{2}(2n - 1 + 2n - 3) = \frac{a}{2}(4n - 4) = 2a(n - 1)$.Substituting $a = \frac{v}{n}$,we get $S_{total} = 2(\frac{v}{n})(n - 1) = \frac{2v(n - 1)}{n}$.

$A$ body starts from rest with uniform acceleration and its velocity at a time of $n$ seconds is $v$. The total displacement of the body in the $n^{\text{th}}$ and $(n-1)^{\text{th}}$ seconds of its motion is:

Similar Questions

When are the acceleration and average acceleration equal for a moving object?

If the velocity of a particle moving along a straight line with uniform acceleration is given by $V = \sqrt{196 - 16x} \text{ m/s}$, then its acceleration is ($x$ is the displacement of the particle). (in $\text{ m/s}^2$)

$A$ particle is moving along the $X$-axis with velocity $v = e^{-\beta x}$. At time $t = 0$,the particle is located at $x = 0$. The displacement of the particle as a function of time is

$A$ body falling for $2 \, s$ covers a distance $S$ equal to that covered in the next second. Taking $g = 10 \, m/s^2$,$S = .......... m$

$A$ 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