A driver parks his car on a hill but accidently leaves it in neutral. The car starts from rest and accelerates down the 45 m long hill. It take the car 12.3 seconds to reach the bottom of the hill.

a) What is the final velocity of the car at the bottom of the hill? (ANS: 7.3 m/s)

b) Once at the bottom of the hill, the car accelerates due to friction at 1.2 m/s2 [backward] along the level ground. How far will the car coast before stopping once it is off the hill? (ANS: 22.3 m)

***show steps please***

a. average velocity=distance/time=45/12.3 m/s

final velocity=2 x average velocity
= 90/12.3 m/s

b. vf^2=vi^2+2ad
0=(90/12.3)^2 + 2(-1.2)d
solve for d

To solve this problem, we can use the equations of motion. Let's break it down into two parts:

Part A: Acceleration down the hill
We can use the equation v = u + at, where:
- v is the final velocity,
- u is the initial velocity (which is 0, as the car starts from rest),
- a is the acceleration (due to gravity, which is approximately 9.8 m/s^2 for objects near the Earth's surface), and
- t is the time taken (which is given as 12.3 seconds).

Using the equation, we can solve for v:

v = u + at
v = 0 + (9.8 m/s^2)(12.3 s)
v = 120.54 m/s

However, since gravity pulls the car downhill, the velocity we obtained is in the opposite direction. Therefore, the final velocity at the bottom of the hill is -120.54 m/s.

Part B: Acceleration due to friction on level ground
We first need to find the time it takes for the car to stop. This can be done using the equation v = u + at, where:
- u is the initial velocity (which is -120.54 m/s),
- a is the acceleration due to friction (which is -1.2 m/s^2, since it acts in the opposite direction), and
- v is the final velocity (which is 0).

Using the equation, we can solve for t:

v = u + at
0 = -120.54 m/s - (1.2 m/s^2)t
120.54 m/s = (1.2 m/s^2)t
t = 100.45 seconds

Now that we have the time, we can calculate the distance traveled using the equation s = ut + (1/2)at^2, where:
- u is the initial velocity (which is -120.54 m/s),
- a is the acceleration due to friction (which is -1.2 m/s^2), and
- t is the time taken (which is 100.45 seconds).

Using the equation, we can solve for s:

s = ut + (1/2)at^2
s = (-120.54 m/s)(100.45 s) - (1/2)(1.2 m/s^2)(100.45 s)^2
s = -12094.07 m - 6028.94 m
s = -18123.01 m

Since the car travels in the negative direction, the distance traveled off the hill is positive. Therefore, the car will coast for approximately 18123.01 meters before stopping.

However, in the question, it is mentioned that the car travels 45 meters down the hill. This means that the distance traveled off the hill will be the difference between the total distance traveled and the distance down the hill:

Distance off the hill = Total distance traveled - Distance down the hill
Distance off the hill = 18123.01 m - 45 m
Distance off the hill = 18078.01 m

So, the car will coast for approximately 18078.01 meters before stopping.

Note: The negative sign in both the velocity and distance values indicates the direction of motion, i.e., backward.