Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.
So now we show it > 
Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.
Which the formula for constant acceleration would be > 
The initial velocity is 50mi/h 
When it stops the final velocity is 
Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27
Then we substitute the values in....

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 
So this means that the car traveled in feet 70.8 ft before it came to a stop.
Answer:
I'm sure it's Nutritional Imbalance
Explanation:
Frictional force and Applied force has same “magnitude” and “opposite” direction.
Option: B
<u>Explanation</u>:
When a book is moved horizontally by applying “force” on the book, the frictional force is opposed to the book by the table. Here, this “frictional force” is opposing the book has the same force what we applied on the book but this frictional force and the applied force are opposite in direction. Always the “frictional force” is opposite to the “applied force” which stops the object to move. For example, if a force applied leftward to the object the frictional force is acted on the right side of the object.
When two objects are in contact they experience a "frictional force". This "frictional force" acts opposite to the force applied on to move the object.
Formula for "frictional force" is 
Where,
is coefficient of friction and N is normal force.