Question

A particle starts moving with acceleration $2m/{s^2}$. Distance travelled by it in ${5^{th}}$ half second is:
a. $1.25m$
b. $2.25m$
c. $6.25m$
d. $30.25m$

Answer 1

Apply the kinematic equation of motion, where you substitute the value of $u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity. Apply the kinematic equation to find the distance in the ${5^{th}}$ half second. Distance in ${5^{th}}$ half second is the difference between the distance travelled in $2s$ and $2.5s$.

Formula Used:
$s = ut + \dfrac{1}{2}g{t^2}$
$u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity.

Complete step by step answer:
Motion equation helps to describe a body’s location, velocity or acceleration relative to frame of reference. The velocity and the position can be derived from the newton equation by the method of integration. Here force acting on a body is known as the function of time.
Kinematics is the study of motion of the given bodies without considering the reason of motion. Length is the path between final and initial position.
Displacement may or may not be equal to the path length travelled of an object. Distance to unit time is called speed. It is a scalar quantity.
A uniform motion along the straight line when that the body is moving with uniform velocity is known as a body.

Distance in ${5^{th}}$ half second is the difference between the distance travelled in $2s$ and $2.5s$.
Distance in ${5^{th}}$ half second = Distance travelled in $2.5s$- Distance travelled in $2s$
$s = ut + \dfrac{1}{2}g{t^2}$
Here the initial velocity is zero.
$s = \dfrac{1}{2}a\left( {t_1^2 - t_2^2} \right)$
Putting the values and we get
$s = 0.5 \times 2 \times \left( {{{2.5}^2} - {2^2}} \right)$
On some simplification we get,
$s = 2.25m$

Hence, the correct answer is option (B).

Note: The initial velocity will be zero if the motion starts from rest and the frame of reference should be the same. The velocity and the position can be derived from the newton equation by the method of integration. The velocity equation integration results in the distance equation.