A moving body is covering distances which are | vedclass

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

Question

(B) Given that the distance $s$ is proportional to the square of the time $t$,we have $s \propto t^{2}$.This can be written as $s = k t^{2}$,where $k$

Vedclass · Accepted Answer

Constant but not Zero

Vedclass · Answer

(B) Given that the distance $s$ is proportional to the square of the time $t$,we have $s \propto t^{2}$.This can be written as $s = k t^{2}$,where $k$ is a constant.The velocity $v$ is the first derivative of distance with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(k t^{2}) = 2kt$.The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$.Since $k$ is a constant,$2k$ is also a constant and is not equal to zero.Therefore,the acceleration of the body is constant but not zero.

$A$ moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is

Similar Questions

Which of the following represents uniformly accelerated motion?

$A$ body starts from rest with uniform acceleration. If its velocity after $n$ seconds is $\upsilon$,then its displacement in the last two seconds is

$A$ particle moves with constant acceleration. Let $v_1, v_2, v_3$ be the average velocities in successive time intervals $t_1, t_2$ and $t_3$. Then:

The initial velocity of a particle is $u$ (at $t = 0$) and the acceleration is given by $a(t) = at$. Which of the following relations is valid?

The velocity of a body moving with a uniform acceleration of $2\,m/s^2$ is $10\,m/s$. Its velocity after an interval of $4\,s$ is...........$m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module