A particle starts its motion from rest under the action of a constant force. If the distance covered in the fi

Question

A particle starts its motion from rest under the action of a constant force. If the distance covered in the first 10 s is S₁ and that covered in the first 20 s is S₂, then:

Accepted Answer

1.  **Identify Conditions:** The particle starts from rest ($u = 0$) under a constant force, which implies constant acceleration ($a$) according to Newton's Second Law ($F=ma$) .
2.  **Select Kinematic Equation:** For a particle moving with constant acceleration starting from rest, the distance ($S$) covered in time ($t$) is given by the equation:
    $$ S = ut + \frac{1}{2}at^2 $$
    Since $u=0$, this simplifies to:
    $$ S = \frac{1}{2}at^2 $$
    Thus, distance is directly proportional to the square of time ($S \propto t^2$) .
3.  **Calculate Distances:**
    *   Distance covered in the first 10 seconds ($t_1 = 10$ s):
        $$ S_1 = \frac{1}{2}a(10)^2 = 50a $$
    *   Distance covered in the first 20 seconds ($t_2 = 20$ s):
        $$ S_2 = \frac{1}{2}a(20)^2 = 200a $$
4.  **Determine Relationship:**
    Taking the ratio of $S_2$ to $S_1$:
    $$ \frac{S_2}{S_1} = \frac{200a}{50a} = 4 $$
    $$ S_2 = 4S_1 $$

Step-by-Step Solution

Practice All PHYSICS Questions