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
Answer:
Step-by-step explanation:
This is an inverse proportion.
men = k / Time So first we need to find k
k = men * Time Substitute values
k = 15*80
k = 1200 Notice the units are man days.
Now what happens if we increase the number of men to 20? What happens to the number of days.
20 = 1200 / time Multiply both sides by time
20*time = 1200 Now divide by 20
time = 1200/20
Answer: time = 60 days
Answer: quotient is 2x^2 + 10x - 5
Solution:
The first polynomial is miswritten.
The right one is: 2x^3 + 4x^2 - 35x + 15.
So, the division is [2x^3 + 4x^2 - 35x + 15] / (x - 3)
The synthetic division uses the coeffcients and obviate the letters, but you have to be sure to respect the place of the coefficient.
So, in this case it is:
3 | 2 4 -35 15
---------------------------------
2 10 - 5 0
So, the quotient is 2x^2 + 10x - 5, and the remainder is 0.
I like to show it in this other way:
| 2 4 -35 15
|
|
3 | +6 +30 -15
--------------------------------
2 10 - 5 0
Of course they are the same coefficients and the answer continue being quotien 2x^2 + 10x - 5, remainder 0.
90+55+75+60+80+90=450
450/6=75
