Answer:
the answer is 2.8
Step-by-step explanation:
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
<span>g(x) = x^2 + 3x - 4 (please be sure to use " ^ " to denote exponentiation).
This function factors into g(x) = (x+4)(x-1), and so the zeros are -4 and +1.
If we shift this graph 5 units to the left, we get h(x) = (x+5)^2 + 3(x+5) - 4.
The new zeros will be -4-5 and 1-5, or -9 and -4.
</span>
I have no idea either sorry