Answer:
a) 4.35 m/s²
b) 2.73 m/s²
c) 7.25 m/s
d) 8.06 m/s
e) At t = 2 s
x = 16.5 m
v = 7.88 m/s
a = 0.099 m/s²
f) t = 0.743 s
Explanation:
Force balance on the rock
ma = 17.4 - F
4a = 17.4 - kv
4a = 17.4 - 2.16v
a) At the initial instant, F = kv = 0
4a = 17.4
a = 4.35 m/s²
b) When v = 3 m/s
4a = 17.4 - (2.16)(3) = 10.92
a = 2.73 m/s²
c) a₀ = 4.35 m/s²
0.1 a₀ = 0.435 m/s²
4a = 17.4 - 2.16v
4(0.435) = 17.4 - 2.16v
1.74 = 17.4 - 2.16v
2.16v = 15.66
v = 7.25 m/s
d) Terminal speed is when the body stops accelerating in the fluid
When a = 0
0 = 17.4 - 2.16v
2.16 v = 17.4
v = 8.06 m/s
e) 4a = 17.4 - 2.16v
a = 4.35 - 0.54 v
But a = dv/dt
(dv/dt) = 4.35 - 0.54v
∫ dv/(4.35 - 0.54v) = ∫ dt
Integrating the left hand side from 0 to v and the right hand side from 0 to t
- 1.852 In (4.35 - 0.54v) = t
In (4.35 - 0.54v) = - 0.54 t
4.35 - 0.54v = e⁻⁰•⁵⁴ᵗ
0.54v = 4.35 - e⁻⁰•⁵⁴ᵗ
v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ
Then, v = dx/dt
(dx/dt) = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ
dx = (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt
∫ dx = ∫ (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt
Integrating the left hand side from 0 to x and the right hand side from 0 to t
x = 8.06t + e⁻⁰•⁵⁴ᵗ
Acceleration too can be obtained as a function of time
since v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ and a = dv/dt
a = 0.54² e⁻⁰•⁵⁴ᵗ = 0.2916 e⁻⁰•⁵⁴ᵗ
At t = 2 s
Coordinate
x = 8.06t + e⁻⁰•⁵⁴ᵗ
x = (8.06)(2) + e^(-1.08) = 16.5 m down into the fluid.
Velocity
v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ
v = 8.06 - 0.54 e^(-1.08) = 7.88 m/s
Acceleration
a = 0.2916 e⁻⁰•⁵⁴ᵗ
a = 0.2916 e^(-1.08) = 0.099 m/s²
f) t = ? When v = 0.9 × 8.06 = 7.254 m/s
v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ
7.254 = 8.06 - 0.54e⁻⁰•⁵⁴ᵗ
- 0.806 = - 0.54 e⁻⁰•⁵⁴ᵗ
e⁻⁰•⁵⁴ᵗ = 1.493
0.54t = In 1.493 = 0.401
t = 0.743 s.
