A particle starting from rest moves with a un | vedclass

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

Question

(C) Let the uniform acceleration of the particle be $a$ and initial velocity $u = 0 \ m/s$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,the

Vedclass · Accepted Answer

$3x \ m$

Vedclass · Answer

(C) Let the uniform acceleration of the particle be $a$ and initial velocity $u = 0 \ m/s$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,the distance covered in the first $5 \ s$ is:$x = 0(5) + \frac{1}{2}a(5)^2 = 12.5a \implies a = \frac{x}{12.5} = 0.08x \ m/s^2$.Now,the distance covered in the first $10 \ s$ (total time) is:$s_{10} = 0(10) + \frac{1}{2}a(10)^2 = 50a$.Substituting $a = 0.08x$:$s_{10} = 50(0.08x) = 4x \ m$.The distance covered in the next $5 \ s$ is the difference between the distance covered in $10 \ s$ and the distance covered in $5 \ s$:$	ext{Distance} = s_{10} - x = 4x - x = 3x \ m$.

$A$ particle starting from rest moves with a uniform acceleration and covers $x$ meters in the first $5 \ s$. The same particle will cover the following distance in the next $5 \ s$:

Similar Questions

$A$ particle moves along the $x$-axis as $x = 4(t-2) + a(t-2)^2$. Which of the following statements is true?

$A$ bullet is fired into a fixed target and loses one-third of its velocity after traveling $4 \ cm$. It penetrates a further $D \times 10^{-3} \ m$ before coming to rest. The value of $D$ is:

$A$ body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ second to that covered in $5$ seconds is

If average and instantaneous velocities are same,then what is the type of motion?

The velocity $u$ and displacement $r$ of a body are related as $u^2 = kr$,where $k$ is a constant. What will be the velocity after $1 \, s$? (Given that the displacement is zero at $t = 0$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module