27. Q A particle moves from rest in a straight line with uniform acceleration. it covered 24 m. in the first four seconds of starting its motion, then the magnitude of its acceleration \( =\ldots \ldots . . . . . . . \mathrm{m} . \mathrm{sec} \). \( \begin{array}{llll}\text { A } 3 & \text { B } 6 & \text { C } 12 & \text { D } \frac{3}{2}\end{array} \)

To find the acceleration of the particle, we can use the kinematic equation that relates distance, initial velocity, time, and acceleration: \[ s = ut + \frac{1}{2} a t^2 \] where: - \( s \) = distance covered (24 m) - \( u \) = initial velocity (0 m/s, since the particle starts from rest) - \( t \) = time (4 s) - \( a \) = acceleration (what we are trying to find) Substituting the known values into the equation: \[ 24 = 0 \cdot 4 + \frac{1}{2} a (4^2) \] \[ 24 = \frac{1}{2} a (16) \] \[ 24 = 8a \] Now, solving for \( a \): \[ a = \frac{24}{8} = 3 \, \text{m/s}^2 \] Thus, the magnitude of the acceleration is \( 3 \, \text{m/s}^2 \). The answer is \( \boxed{3} \).