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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA particle moves along the x-axis such that its position is given by \( x = a + b t^2 \), where \( a = 8.5 \text{ m} \) and \( b = 2.5 \text{ m/s}^2 \). A second particle moves along the same line starting from rest at origin with constant acceleration \( c \). If the average velocity of the first particle between \( t = 2.0 \text{ s} \) and \( t = 4.0 \text{ s} \) equals the instantaneous velocity of the second particle at \( t = 3.0 \text{ s} \), find the value of \( c \).
Correct answer · Option 3
\( 5.0 \text{ m/s}^2 \)
Explanation
For the first particle, average velocity \( v_{\text{avg}} = \frac{x(4) - x(2)}{4 - 2} = \frac{[8.5 + 2.5(16)] - [8.5 + 2.5(4)]}{2} = \frac{48.5 - 18.5}{2} = 15 \text{ m/s} \). For the second particle, \( v_2(t) = c t \implies v_2(3) = 3 c \). Equating gives \( 3 c = 15 \implies c = 5.0 \text{ m/s}^2 \).
