We are asked to find the rate of acceleration of an object, given its initial velocity, distance traveled, and time of travel. The object is accelerating at a constant rate, so we can use the equations of motion for constant acceleration.

The equation that relates distance traveled, initial velocity, time, and acceleration is:

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

Where:

• $d = 998 \, \text{ft}$ (distance traveled),
• $v_0 = 14.7 \, \text{ft/sec}$ (initial velocity),
• $t = 15.0 \, \text{seconds}$ (time),
• $a$ is the acceleration we need to find.

Now, substitute the known values into the equation:

$998 = 14.7(15.0) + \frac{1}{2} a (15.0)^2$

Simplify the terms:

$998 = 220.5 + \frac{1}{2} a (225)$

Next, isolate the acceleration term $a$:

$998 - 220.5 = \frac{1}{2} a (225)$

$777.5 = \frac{1}{2} a (225)$

Multiply both sides by 2 to eliminate the fraction:

$1555 = a \times 225$

Now, solve for $a$:

$a = \frac{1555}{225}$

$a \approx 6.91 \, \text{ft/sec}^2$

Thus, the magnitude of the acceleration is approximately $6.91 \, \text{ft/sec}^2$.

Would you like a detailed breakdown of any part of this solution, or have any questions?

Here are 5 related questions to explore further:

1. How would the result change if the object were initially at rest?
2. What is the final velocity of the object after 15 seconds?
3. How much time would the object take to travel 2000 ft with the same acceleration?
4. How would the calculation change if the time interval were 20 seconds instead of 15?
5. Can this equation be applied if the object decelerates instead of accelerating?

Tip: Always check if the initial velocity is zero or non-zero, as it significantly affects the equation used for motion analysis.