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

Question

A particle starts its motion from rest under the action of a constant force. If the distance covered in 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 ($u = 0$) and moves under a constant force, which implies constant acceleration ($a$).
2.  **Select Kinematic Equation:** The distance covered ($s$) in time ($t$) under constant acceleration is given by the second equation of motion:
    $$ s = ut + \frac{1}{2}at^2 $$ 
    Since $u = 0$, the equation 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:**
    *   For the first 10 seconds ($t_1 = 10$ s):
        $$ s_1 = \frac{1}{2}a(10)^2 = 50a $$
    *   For the first 20 seconds ($t_2 = 20$ s):
        $$ s_2 = \frac{1}{2}a(20)^2 = 200a $$
4.  **Find Relationship:**
    Divide $s_2$ by $s_1$:
    $$ \frac{s_2}{s_1} = \frac{200a}{50a} = 4 $$
    $$ s_2 = 4s_1 $$

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