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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA particle moving along a straight line under constant deceleration covers a distance of \( 20 \text{ m} \) in the \( 3^{\text{rd}} \) second and \( 12 \text{ m} \) in the \( 5^{\text{th}} \) second. Determine the total distance covered by the particle before it comes to rest.
Correct answer · Option 1
\( 112.5 \text{ m} \)
Explanation
Distance in \( n^{\text{th}} \) second: \( D_n = v_0 + \frac{a}{2}(2n-1) \). So \( D_3 = v_0 + 2.5a = 20 \) and \( D_5 = v_0 + 4.5a = 12 \). Subtracting gives \( 2a = -8 \implies a = -4 \text{ m/s}^2 \). Substituting gives \( v_0 = 30 \text{ m/s} \). Total distance before coming to rest: \( S = \frac{v_0^2}{2|a|} = \frac{30^2}{2 \times 4} = 112.5 \text{ m} \).
