Question

Pls someone should help me answer question 3. Thank you​

Answer 1

(a) 12 m/s

(b) At t = 0, x'(t) = 0

At t = 5.0 s, x'(t) = 15 m/s

At t = 10.0 s, x'(t) = 12 m/s

(c) i. The initial velocity = 2.0 cm/s

The initial position = 50 cm

The initial acceleration = 0.125 cm/s²

ii. 16 seconds

2. (a) Approximately 0.85 m/s

(b) 2.094 J

(c) i. Approximately 49.74 m/s

ii. Approximately 86.38°

3. (i) 4·i + 5·j

(ii) -2·i - j - 2·k

(iii) 5·i + 4·j - 3·z

(iv) 8

(v) (a) The magnitude is 2.8 cm, the direction is East

(b) The magnitude is ((14·√3)/5) cm, the direction is North

(c) The magnitude is ((14·√3)/5), the direction is South

Step-by-step explanation:

x(t) is given as follows;

x(t) = b·t² - c·t³

Where;

b = 2.4 m/s(²) (we note that the unit of b for the term to be distance in m/s²)

c = 0.120 m/s³

(a) At t = 0, we have;

x(0) = b × 0² - c × 0³ = 0

At t = 10.0 s, we have;

x(10.0) = 2.4 m/s² × (10.0 s)² - 0.120 m/s³ × (10.0 s)³ = 120 m

The average velocity = (Total distance)/(Total time)

∴ The average velocity of the car for the time interval t = 0 to t = 10.0 s,
`v_(ave)`, is given as follows;

`v_(ave)` = (120 m - 0 m)/(10.0 s - 0 s) = 12 m/s

The average velocity of the car for the time interval t = 0 to t = 10.0 s,
`v_(ave)` = 12 m/s

(b) The instantaneous velocity,
`v_(inst)`, is given as follows;

`v_(inst) = \lim \limits_(t \to 0) \left( v_(ave)\right) = \lim \limits_(t \to 0) \left( (\Delta x)/(\Delta t) \right) = (dx)/(dt) = x'(t)`

`x'(t) = (d\left (b \cdot t^2 - c \cdot t^3\right))/(dt)`

x'(t) = 2·b·t - 3·c·t²

At t = 0, x'(t) = 2 × b × 0 - 3 × c × 0² = 0

At t = 5.0 s, x'(t) = 2 × 2.4 m/s² × 5.0 s - 3 × 0.120 m/s³ × (5.0 s)² = 15 m/s

At t = 10.0 s, x'(t) = 2 × 2.4 m/s² × 10.0 s - 3 × 0.120 m/s³ × (10.0 s)² = 12 m/s

(c) x(t) = 50 cm + (2.0 cm/s)·t - (0.0625 cm/s²)·t²

i. The initial velocity is the instantaneous velocity, x'(t), at time, t = 0

x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t

At t = 0, x'(0) = 2.0 cm/s - 2 × 0.0625 cm/s² × 0 = 2.0 cm/s

The initial velocity, x'(0) = 2.0 cm/s

The initial position = The position at time t = 0 = x(0)

x(0) = 50 cm + (2.0 cm/s) × 0 - (0.0625 cm/s²) × 0² = 50 cm

The initial position, x(0) = 50 cm

The initial acceleration, x''(0) = 2 × 0.0625 cm/s² = 0.125 cm/s²

ii. x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t

When the velocity of the turtle, x'(t) = 0 we have;

0 = 2.0 cm/s - 2 × 0.0625 cm/s² × t

∴ t = (2.0 cm/s)/(2 × 0.0625 cm/s²) = 16 seconds

The velocity of the turtle is zero after 16 seconds

2. The mass of the large fish, m₁ = 15.0-kg

The speed of the large fish, v₁ = 1.1 m/s

The mass of the smaller fish, m₂ = 4.50 kg

The speed of the small (stationary) fish, v₂ = 0

The initial momentum = 15.0 kg × 1.1 m/s + 4.50 kg × 0 = 16.5 kg·m/s

The initial momentum = 16.5 kg·m/s

The final momentum = (15.0 kg + 4.50 kg) × v₃ = 19.50 kg × v₃

The final momentum = 19.50 kg × v₃

Where;

The total initial momentum = The total final momentum

We get;

16.5 kg·m/s = 19.50 kg × v₃

∴ v₃ = (16.5 kg·m/s)/(19.50 kg)

v₃ = (16.5/19.50) m/s = (11/13) m/s ≈ 0.85 m/s

∴ The speed of the large fish just after it eats the small, v₃ ≈ 0.85 m/s

(b) The initial kinetic energy, K.E.₁ = (1/2) × 15 kg × (1.1 m/s)² = 9.075 J

The final kinetic energy, K.E.₂ = (1/2) × 19.5 kg × (11/13 m/s)² = 363/52 J

The mechanical energy dissipated, ΔE = K.E.₁ - K.E.₂

ΔE = (9.075 - 363/42) J = 1089/520 J ≈ 2.094 J

The mechanical energy dissipated, ΔE = 2.094 J

(c) i. We have the total momentum = 110 × 8.8· j + 85 × 7.2· i = 9.680·i + 612·i

The velocity after collision, v = (9.680·i + 612·i)/(110 + 85) = 49.64·j + 3.14·i

The magnitude of the velocity, v = √(49.64² + 3.14²) ≈ 49.74 m/s

ii. The direction, θ = arctan(49.64/3.14) ≈ 86.38°

3. (i)
`\underset{A}{\rightarrow} + \underset{B}{\rightarrow}` = (i + 2·j - k) + 3·i + 3·j + k = 4·i + 5·j

(ii)
`\underset{A}{\rightarrow} - \underset{B}{\rightarrow}` = (i + 2·j - k) - (3·i + 3·j + k) = -2·i - j - 2·k

(iii)
`\underset{A}{\rightarrow} * \underset{B}{\rightarrow}` = (2 + 3)·i - (1 + 3)·j + (3 - 6)·z = 5·i + 4·j - 3·z

(iv)
`\underset{A}{\rightarrow} \cdot \underset{B}{\rightarrow}` = 1×3 + 2 × 3 + (-1)×1 = 8

(v) (a)
`\underset{A}{\rightarrow} + \underset{B}{\rightarrow}` = 2.8·cos(60°)·i + 2.8·sin(60°)·j + (2.8·cos(60°)·i - 2.8·sin(60°)·j

`\underset{A}{\rightarrow} + \underset{B}{\rightarrow}` = 5.6·cos 60°·i = 2.8·i

The magnitude = 2.8, the direction is east

(b)
`\underset{A}{\rightarrow} - \underset{B}{\rightarrow}` = 2.8·cos(60°)·i + 2.8·sin(60°)·j - (2.8·cos(60°)·i - 2.8·sin(60°)·j

`\underset{A}{\rightarrow} - \underset{B}{\rightarrow}` = 5.6·sin(60°)·j= ((14·√3)/5)·j

The magnitude = ((14·√3)/5), the direction is North

(c)
`\underset{B}{\rightarrow} - \underset{A}{\rightarrow}` = (2.8·cos(60°)·i - 2.8·sin(60°)·j - (2.8·cos(60°)·i + 2.8·sin(60°)·j)

`\underset{B}{\rightarrow} - \underset{A}{\rightarrow}` = -5.6·sin(60°)·j= (-(14·√3)/5)·j

The magnitude = ((14·√3)/5), the direction is South