This is the concept of algebra; given that the height of the house at time t is given by:
h(t)=<span> 400 − 39t − 157e−t/4
the velocity of the house is given by:
dh/dt
velocity of our object is:
dh/dt=-39+1/4*157e^-(t/4)
=-39+157/4 e^(-t/4)
thus the velocity at h=0 will be as follows:
at h=0, the value of t will be:
0=400-39t-157e^(-t/4)
solving for t we get:
t=-5.4 or t=10.1
since we don't have negative time, t=10.1
thus the velocity at this point will be:
dh/dt=-38+1/4*157e^(-10.1/4)
=-38.86 sec
this means that the object dropped by a velocity of 38.86 sec</span>
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>
Answer:
y=-2x+6 and y=4x
Step-by-step explanation:
If you graph these two, the point where the two lines intersect is (1,4).
706 ÷ 0363
1.94490358127 is your answer.
