Final answer:
To stop a car traveling at 120 mi/h in 200 ft, we first convert the speed to ft/s, use the kinematic equation to solve for acceleration in ft/s^2, and then convert it back to mi/h^2, resulting in an acceleration of approximately -6000 mi/h^2.
Step-by-step explanation:
To calculate the acceleration needed for a car traveling at 120 mi/h to stop in 200 ft, we must first convert the speed to feet per second (ft/s). Since 1 mile = 5280 feet and 1 hour = 3600 seconds, the conversion is as follows:
120 mi/h = 120 * (5280 ft/mi) / (3600 s/h) ≈ 176.00 ft/s.
Using the kinematic equation v2 = u2 + 2as, where v is the final velocity (0 ft/s), u is the initial velocity (176.00 ft/s), a is the acceleration, and s is the displacement (200 ft), we have:
0 = (176.00)2 + 2 * a * 200. To find acceleration, a, we rearrange the equation:
a = - (176.00)2 / (2 * 200) ≈ -154.88 ft/s2.
The negative sign indicates deceleration. Now, we convert the acceleration back to miles per hour squared (mi/h2):
a = -154.88 ft/s2 * (3600 s/h)2 / (5280 ft/mi). After calculation, this gives us:
a ≈ -6042 mi/h2 (to two significant figures, a ≈ -6000 mi/h2).
The acceleration required to stop the car in 200 ft while traveling at 120 mi/h is approximately -6000 mi/h2.