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 $$ =\ldots \ldots \ldots . . $$. m.sec.

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To find the acceleration of the particle, we can use the kinematic equation for uniformly accelerated motion:

$$
s = ut + \frac{1}{2} a t^2
$$

where:
- $$ s $$ is the distance covered (24 m),
- $$ u $$ is the initial velocity (0 m/s, since the particle starts from rest),
- $$ a $$ is the acceleration (which we need to find),
- $$ t $$ is the time (4 seconds).

Since the particle starts from rest, we can substitute $$ u = 0 $$ into the equation:

$$
s = 0 \cdot t + \frac{1}{2} a t^2
$$

This simplifies to:

$$
s = \frac{1}{2} a t^2
$$

Now, substituting the known values $$ s = 24 $$ m and $$ t = 4 $$ s into the equation:

$$
24 = \frac{1}{2} a (4^2)
$$

Calculating $$ 4^2 $$:

$$
4^2 = 16
$$

Now substituting this back into the equation:

$$
24 = \frac{1}{2} a \cdot 16
$$

This simplifies to:

$$
24 = 8a
$$

Now, solving for $$ a $$:

$$
a = \frac{24}{8} = 3 \, \text{m/s}^2
$$

Thus, the magnitude of the acceleration is:

$$
\boxed{3} \, \text{m/s}^2
$$
The acceleration is 3 m/s².

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 \( =\ldots \ldots \ldots . . \). m.sec.

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