First let us find


Now let;s solve the second part
Fern's estimate is not correct because he rounded 548 wrong, it would be rounded to 550 instead since the number 8 is closer to 10. So the correct answer would be 420.
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
Here is your answer and an explanation
Answer: y = -2/3x
Explanation:
This can be determined by calculating the gradient of the straight line, using:
m=ΔyΔx
=−6−34−(−2)
=−96
=−32
Then we use the slope-point form of the straight line:
y−y1=m(x−x1)
to give:
y−3=−32(x−(−2))
∴y−3=−32(x+2)
∴y−3=−32x−3
∴y=−32x