If the Khans car runs 22 miles per one gallon, then on 5 gallons of gas they can drive 110 miles.
1 gallon = 22 miles
5g x 22m = 110
I hope this helps! If I did something wrong, please correct me!
Part A
Given that

Then,

For

, then

Thus,

For

, we have

Part B
Recall that from part A,

Now, at initial position, t = 0 and

, thus we have

and when the velocity drops to half its value,

and

Thus,

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by
f(x) = 5x + 1
f(4) = 5 * 4 + 1
f(4) = 20 + 1
f(4) = 21
g(x) = 2x² - 3x +1
g(5) = 5 * 5² - 3 * 5 + 1
g(5) = 125 - 15 + 1
g(5) = 111
f(4) + g(5) = 21 + 111 = 132