8) $$ f(t)=-4 \sin 2 t $$, and $$ v(0)=2, x(0)=-3 $$, then $$ x(\pi)= $$ (a) -3 (b) zero (c) 2 (d) 3 9) If $$ \mathrm{v}=3 \mathrm{t}-2 $$, then s in the interval $$ [0,2] $$ is $$ \qquad $$ unit length. (a) 1 (b) 2 (c) 3 (d) 4 10)If $$ v=3 t^{2}-2 t $$, then the distance covered within the interval $$ [0,2] $$ $$ \qquad $$ unit length. (a) $$ \frac{4}{27} $$ (b) 4 (c) $$ \frac{112}{27} $$ (d) $$ \frac{116}{27} $$ 11) If $$ \mathrm{v}=\mathrm{t}^{3}-3 \mathrm{t}^{2}+2 t $$, then the distance covered within the time interval $$ [0,3] $$ $$ \qquad $$ unit length. (a) $$ \frac{1}{4} $$ (b) $$ \frac{1}{2} $$ (c) $$ \frac{9}{4} $$ (d) $$ \frac{11}{4} $$ 12)If $$ a=3, v_{0}=-1 $$, then $$ s $$ within the time interval $$ [0,2] $$ $$ \qquad $$ unit lengh (a) $$ \frac{1}{6} $$ (b) 4 (C) $$ \frac{25}{6} $$ (d) $$ \frac{13}{3} $$ 13)fra=3,$$ v_{0}=-1 $$, then the distance covered within the time interval $$ [0,2] $$ unit length. (a) $$ \frac{1}{6} $$ (b) 4 (C) $$ \frac{25}{6} $$ (d) $$ \frac{13}{3} $$

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### Problem 8
Given $$ f(t) = -4 \sin(2t) $$, we need to find $$ x(\pi) $$ given $$ v(0) = 2 $$ and $$ x(0) = -3 $$.

1. **Find the velocity function**: The velocity $$ v(t) $$ is the derivative of the position function $$ x(t) $$. Since $$ f(t) $$ is given, we can assume $$ v(t) = f(t) $$.
   $$
   v(t) = -4 \sin(2t)
   $$

2. **Integrate to find the position function**:
   $$
   x(t) = \int v(t) \, dt = \int -4 \sin(2t) \, dt
   $$
   Using the integral of sine:
   $$
   x(t) = 2 \cos(2t) + C
   $$

3. **Use the initial condition $$ x(0) = -3 $$** to find $$ C $$:
   $$
   x(0) = 2 \cos(0) + C = 2 + C = -3 \implies C = -5
   $$
   Thus, the position function is:
   $$
   x(t) = 2 \cos(2t) - 5
   $$

4. **Evaluate $$ x(\pi) $$**:
   $$
   x(\pi) = 2 \cos(2\pi) - 5 = 2(1) - 5 = -3
   $$

**Answer for Problem 8**: (a) -3

---

### Problem 9
Given $$ v = 3t - 2 $$, we need to find the distance $$ s $$ in the interval $$ [0, 2] $$.

1. **Integrate the velocity function**:
   $$
   s = \int_0^2 v(t) \, dt = \int_0^2 (3t - 2) \, dt
   $$

2. **Calculate the integral**:
   $$
   s = \left[ \frac{3t^2}{2} - 2t \right]_0^2 = \left( \frac{3(2^2)}{2} - 2(2) \right) - \left( 0 \right) = \left( \frac{12}{2} - 4 \right) = 6 - 4 = 2
   $$

**Answer for Problem 9**: (b) 2

---

### Problem 10
Given $$ v = 3t^2 - 2t $$, we need to find the distance covered within the interval $$ [0, 2] $$.

1. **Integrate the velocity function**:
   $$
   s = \int_0^2 (3t^2 - 2t) \, dt
   $$

2. **Calculate the integral**:
   $$
   s = \left[ t^3 - t^2 \right]_0^2 = \left( 2^3 - 2^2 \right) - \left( 0 \right) = 8 - 4 = 4
   $$

**Answer for Problem 10**: (b) 4

---

### Problem 11
Given $$ v = t^3 - 3t^2 + 2t $$, we need to find the distance covered within the time interval $$ [0, 3] $$.

1. **Integrate the velocity function**:
   $$
   s = \int_0^3 (t^3 - 3t^2 + 2t) \, dt
   $$

2. **Calculate the integral**:
   $$
   s = \left[ \frac{t^4}{4} - t^3 + t^2 \right]_0^3 = \left( \frac{3^4}{4} - 3^3 + 3^2 \right) - \left( 0 \right)
   $$
   $$
   = \left( \frac{81}{4} - 27 + 9 \right) = \left( \frac{81}{4} - \frac{108}{4} + \frac{36}{4} \right) = \frac{81 - 108 + 36}{4} = \frac{9}{4}
   $$

**Answer for Problem 11**: (c) $$ \frac{9}{4} $$

---

### Problem 12
Given $$ a = 3 $$ and $$ v_0 = -1 $$, we need to find $$ s $$ within the time interval $$ [0, 2] $$.

1. **Use the formula for distance**:
   $$
   s = v_0 t + \frac{1}{2} a t^2
   $$
   Substituting $$ v_0 = -1 $$, $$ a = 3 $$, and $$ t = 2 $$:
   $$
   s = -1(2) + \frac{1}{2}(3)(2^2) = -2 + \frac{1}{2}(3)(4) = -2 + 6 = 4
   $$

**Answer for Problem 12**: (b) 4

---

### Problem 13
This problem is identical to Problem 12, so the answer remains the same.

**Answer for Problem 13**: (b) 4

---

In summary, the answers are:
- Problem 8: (a) -3
- Problem 9: (b) 2
- Problem 10: (b) 4
- Problem 11: (c) $$ \frac{9}{4} $$
- Problem 12: (b) 4
- Problem 13: (b) 4
- Problem 8: (a) -3 - Problem 9: (b) 2 - Problem 10: (b) 4 - Problem 11: (c) $$ \frac{9}{4} $$ - Problem 12: (b) 4 - Problem 13: (b) 4

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