Question

For a body starting from rest and moving with uniform acceleration, the ratio of the distances covered for $1s$ , $2s$ , $3s$ is

Answer 1

Hint: Use the second equation of motion which gives the relation for distance, initial speed, uniform acceleration, and time for an object in motion. Keep the values of time and find the respective distance to find the ratio of the distance covered.
Formula used:
$S = ut + \dfrac{1}{2}a{t^2}$
Where, $S$ is the distance, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Complete answer:
The standard equations of motion, often known as the laws of constant acceleration, are derived in the case of motion with uniform or constant acceleration. These equations, which contain the constants displacement($S$), velocity (initial and final), time($t$), and acceleration ($a$), govern the motion of a particle.
Using newton’s second equation of motion
$S = ut + \dfrac{1}{2}a{t^2}$
Where, $S$ is the distance, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.
Since the body is starting from rest therefore $u$ will be zero and acceleration is an uniform acceleration so the distance covered by the body will be directly proportional to the square of the time
$S \propto {t^2}$
Therefore, for $1s$ distance covered will be
${S_1} = 1m$
For $2s$ distance covered will be
${S_2} = 4m$
For $3s$ distance covered will be
${S_3} = 9m$
Hence the ratio of the distance covered is $1:4:9$

Note:
Equations of motion are physics equations that describe a physical system's behavior in terms of its motion as a function of time. When a body's acceleration is constant and motion occurs in a straight line, the equation of motion is utilized.