Understanding the Speed-Time Relationship

This problem involves the relationship between speed, distance, and time. When the distance is constant, speed and time are inversely proportional. This means if the speed decreases, the time taken increases proportionally.

Calculating the Usual Time

Let the usual speed of the man be $S$ and his usual time to cover the distance be $T$ hours. The distance covered is $D = S \times T$.

According to the question, the man walks at $\frac{3}{4}$ of his usual speed. So, the new speed is $\frac{3}{4}S$.

Let the time taken at this new speed be $T'$ hours. The distance covered is the same, so $D = (\frac{3}{4}S) \times T'$.

Equating the distances:

$S \times T = (\frac{3}{4}S) \times T'$

We can cancel $S$ from both sides (assuming $S \neq 0$):

$T = \frac{3}{4}T'$

This implies the new time $T'$ is:

$T' = \frac{4}{3}T$

The problem states that the man is $1\frac{1}{2}$ hours late when walking at the reduced speed. This means the new time $T'$ is $1\frac{1}{2}$ hours more than the usual time $T$.

$T' = T + 1\frac{1}{2}$

Now, substitute $T' = \frac{4}{3}T$ into the equation:

$\frac{4}{3}T = T + 1\frac{1}{2}$

Convert the mixed fraction $1\frac{1}{2}$ to an improper fraction:

$1\frac{1}{2} = \frac{(1 \times 2) + 1}{2} = \frac{3}{2}$

The equation becomes:

$\frac{4}{3}T = T + \frac{3}{2}$

To solve for $T$, subtract $T$ from both sides:

$\frac{4}{3}T - T = \frac{3}{2}$

Combine the terms with $T$:

$(\frac{4}{3} - 1)T = \frac{3}{2}$

$(\frac{4}{3} - \frac{3}{3})T = \frac{3}{2}$

$\frac{1}{3}T = \frac{3}{2}$

Multiply both sides by 3 to find $T$:

$T = 3 \times \frac{3}{2}$

$T = \frac{9}{2}$

Convert the improper fraction $\frac{9}{2}$ back to a mixed number:

$T = 4\frac{1}{2}$

Therefore, the man's usual time to cover the distance is $4\frac{1}{2}$ hours.