Final answer:
The car's acceleration is approximately -2.59 m/s² and it takes approximately 10.35 seconds to bring the car to a stop from an initial speed of 60 mi/h over a distance of 456 ft, after converting to SI units.
Step-by-step explanation:
To solve for the car’s acceleration and the time it takes to bring the car to a stop, we first need to convert the given units to SI units. The initial velocity (vi) of the car is 60 mi/h, which is approximately 26.82 m/s. The stopping distance (d) is 456 ft, which is about 139.0 meters.
Using the kinematic equation
vf2 = vi2 + 2ad, where vf is final velocity, vi is initial velocity, a is acceleration, and d is distance, and knowing that vf is 0 m/s since the car comes to a stop, we can rearrange to find acceleration: a = -vi2 / (2d). Substituting the given values, we find a = -26.822 / (2 × 139.0). The negative sign indicates deceleration. After calculating, we find the car’s acceleration to be approximately -2.59 m/s2.
To find the time (t), we use the kinematic equation vf = vi + at. With vf = 0 m/s and the values for vi and a already known, we rearrange to get t = -vi / a. Upon substitution, we get t = -26.82 / (-2.59), which yields a stopping time of approximately 10.35 seconds.