To find out how far the automobile has traveled during the time of acceleration, we can use the following kinematic equation for uniformly accelerated motion:

$d = v_0 t + \frac{1}{2} a t^2$

Where:

• $d$ is the distance traveled.
• $v_0$ is the initial velocity (which is 0 since the car started from rest).
• $a$ is the acceleration (25 ft/sec²).
• $t$ is the time of acceleration (4.56 seconds).

Since the car started from rest, $v_0 = 0$. Therefore, the equation simplifies to:

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

Substituting the given values:

$d = \frac{1}{2} \times 25 \times (4.56)^2$

Now, let's calculate this:

$d = \frac{1}{2} \times 25 \times 20.7936$

$d = 12.5 \times 20.7936 = 259.92 \text{ feet}$

So, the automobile has traveled 259.92 feet during this time of acceleration.

Would you like more details or have any questions?

Here are five related questions you might find interesting:

1. How would the distance change if the time of acceleration was doubled?
2. What would the distance be if the car started with an initial velocity?
3. How does the acceleration affect the distance traveled if it were reduced by half?
4. Can you calculate the velocity of the car at the end of the 4.56 seconds?
5. What if the car continued to accelerate for an additional 5 seconds; how far would it travel then?

Tip: When dealing with acceleration, it's essential to remember that both time and acceleration have a squared relationship to distance.