Answer:
a. v(t)= -6.78
+ 16.33 b. 16.33 m/s
Step-by-step explanation:
The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=
=
. We now multiply both sides of the equation by the integrating factor.
μv' + μkv = μg ⇒ v' + kv = g ⇒ [v]' = g. Integrating, we have
∫ [v]' = ∫g
v = 
+ c
v(t)= + c
.
From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have
9.55 = 9.8 × 15/9 + c
= 16.33 + c
c = 9.55 -16.33 = -6.78.
So, v(t)= 16.33 - 6.78. m/s = - 6.78 + 16.33 m/s
b. Velocity of object at time t = 0.5
At t = 0.5, v = - 6.78
+ 16.33 m/s = 16.328 m/s ≅ 16.33 m/s
A direct variation suggest that the value of x in the equation would greatly affect the value of y such that when x is increasing, y also increases and the other way around. The equation for a direct variation is that,
y = kx
Substituting the given values in the ordered pair,
5 = k(4) ; k = 5/4
<h3>
Answer: log(2)</h3>
Explanation:
The rule to use here is log(x)-log(y) = log(x/y)
So,
log(6)-log(3) = log(6/3) = log(2)
Answer:
Step-by-step explanation:
The standard form of an equation for a straight line is y=mx+b, where m is the slope and b is the y-intercept (the value of y when x = 0).
We can calculate the slope from the two given points, (6,-3) and (-6,-5). Slope is Rise/Run, where Rise is the change in y and Run is the change in x.
From the two given points, starting at (-6,-5) and going to (6,-3):
Rise = (-3 - (-5)) = +2
Run = (6 - (-6)) = 12
Rise/Run (slope) = 2/12 or 1/6
The equation becomes y = (1/6)x + b
We can find b by enterieng either of the two given points and solving for b. I'll pick (6,-3):
y = (1/6)x + b
-3 = (1/6)*(6) + b
-3 = 1 + b [Now you can see why I chose (6,-3)]
b = -4
The equation is y = (1/6)x - 4
Check this with a DESMOS graph (attached).
