NICET - Highway Construction Inspection Practice Exam

In a scenario where a car starts from rest and accelerates at 8 ft/sec² for 6 seconds, what is the final position of the car?

1. 72 ft

2. 96 ft

3. 144 ft

4. 192 ft

The correct answer is: 144 ft

To determine the final position of a car that starts from rest and accelerates at a constant rate, we can use the equation for the position of an object under uniform acceleration. The formula is: \[ s = ut + \frac{1}{2}at^2 \] where: - \( s \) is the final position, - \( u \) is the initial velocity (which is 0 since the car starts from rest), - \( a \) is the acceleration (8 ft/sec² in this case), - \( t \) is the time (6 seconds). Since the car starts from rest, the initial velocity \( u \) equals 0. Plugging the values into the equation: \[ s = 0 \cdot 6 + \frac{1}{2} \cdot 8 \cdot (6^2) \] \[ s = 0 + \frac{1}{2} \cdot 8 \cdot 36 \] \[ s = 4 \cdot 36 \] \[ s = 144 \text{ ft} \] Thus, the final position of the car after 6 seconds of acceleration at 8 ft/sec² is indeed 144