Answer:
10%
Step-by-step explanation:
Answer:
Remember that:
Speed = distance/time.
Then we can calculate the average speed in any segment,
Let's make a model where the average speed at t = t0 can be calculated as:
AS(t0) = (y(b) - y(a))/(b - a)
Where b is the next value of t0, and a is the previous value of t0. This is because t0 is the middle point in this segment.
Then:
if t0 = 100s
AS(100s) = (400ft - 0ft)/(200s - 0s) = 2ft/s
if t0 = 200s
AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s
if t0 = 300s
AS(300s) = (3200ft - 400ft)/(400s - 200s) = 14ft/s
if t0 = 400s
AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s
So for the given options, t = 400s is the one where the velocity seems to be the biggest.
And this has a lot of sense, because while the distance between the values of time is constant (is always 100 seconds) we can see that the difference between consecutive values of y(t) is increasing.
Then we can conclude that the rocket is accelerating upwards, then as larger is the value of t, bigger will be the average velocity at that point.
f(x+1) = 16(x+1)² - (x+1) = 16(x²+2x+1) - x - 1
= 16x² + 32x + 16 - x - 1 = 16x²+31x+15
so
f(x+1)-f(x) = 16x²+31x+15 - (16x² - x)
= 16x²+31x+15 - 16x² + x
= 32x + 15
