Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

s₂ = 2s₁

s₂ = 3s₁

s₂ = 4s₁

s₂ = s₁

Step-by-Step Solution

Identify Conditions: The particle starts from rest ( $u = 0$ ) and moves under a constant force, which implies constant acceleration ( $a$ ).
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 ( $s \propto t^2$ ).
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): $s_2 = \frac{1}{2}a(20)^2 = 200a$
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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