wheres the photo so we can answer and graph????
The answer would be A. I just did this yesterday. Do you want the steps?
Subtraction doesn't mix with mult. or div. Are you sure you've copied this problem down correctly?
"decrease an amount (x) by 19" would be x - 19 (NO multiplication here)
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' + k
v = 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