Part A. To solve for the distance travelled during the
interval, all we have to do is to plug in values of t = 3 and t = 3.5 in the
equation and the difference would be the answer:
when t = 3: s = 16 (3)^2 = 144 m
when t = 3.5: s = 16 (3.5)^2 = 196 m
Therefore the distance travelled within the interval is:
196 m – 144 m = 52 m
<span>Part B. The velocity is calculated by taking the 1st
derivative of the equation. v = ds / dt</span>
s = 16 t^2
ds / dt = 32 t = v
when t = 3: v = 32 (3) = 96 m / s
when t = 3.5: v = 32 (3.5) = 112 m / s
Therefore the average velocity is:
(96 + 112) /2 = 104 m / s
Part C. We can still use the formula v = 32 t and plug in
the value of t = 3
v = 32 t = 32 (3)
v = 96 m / s
<span> </span>
Two the absolute value is however many numbers away from zero any number is.
B is the answer i had this on my test the other day
Answer:
try this
Step-by-step explanation:
desmos graphing will give u answers
