Evaluate integral12 (3x3 + 2x2 - x + 1)\,dx.

$$\int (3x^3 + 2x^2 - x + 1)\,dx = \frac{3x^{3+1}}{3+1} + \frac{2x^{2+1}}{2+1} - \frac{x^{1+1}}{1+1} + x + C$$ $$= \frac{3}{4}x^4 + \frac{2}{3}x^3 - \frac{1}{2}x^2 + x + C$$

$$\left[ \frac{3}{4}x^4 + \frac{2}{3}x^3 - \frac{1}{2}x^2 + x \right]_{1}^{2}$$

First, evaluate at the upper limit $x=2$:

$$F(2) = \frac{3}{4}(2)^4 + \frac{2}{3}(2)^3 - \frac{1}{2}(2)^2 + 2$$ $$F(2) = \frac{3}{4}(16) + \frac{2}{3}(8) - \frac{1}{2}(4) + 2$$ $$F(2) = 12 + \frac{16}{3} - 2 + 2$$ $$F(2) = 12 + \frac{16}{3} = \frac{36}{3} + \frac{16}{3} = \frac{52}{3}$$

Next, evaluate at the lower limit $x=1$:

$$F(1) = \frac{3}{4}(1)^4 + \frac{2}{3}(1)^3 - \frac{1}{2}(1)^2 + 1$$ $$F(1) = \frac{3}{4} + \frac{2}{3} - \frac{1}{2} + 1$$

Find a common denominator, which is 12:

$$F(1) = \frac{9}{12} + \frac{8}{12} - \frac{6}{12} + \frac{12}{12}$$ $$F(1) = \frac{9 + 8 - 6 + 12}{12} = \frac{23}{12}$$

$$\int_{1}^{2} (3x^3 + 2x^2 - x + 1)\,dx = F(2) - F(1) = \frac{52}{3} - \frac{23}{12}$$ $$= \frac{52 \times 4}{3 \times 4} - \frac{23}{12} = \frac{208}{12} - \frac{23}{12}$$ $$= \frac{208 - 23}{12} = \frac{185}{12}$$

The final answer is $\boxed{\frac{185}{12}}$.

Given:
Mass ($m$) = $1600 \text{kg}$
Initial velocity ($u$) = $0 \text{m/s}$ (starts from rest)
Final velocity ($v$) = $18 \text{m/s}$
Time ($t$) = $10 \text{s}$
Frictional force ($F_f$) = $500 \text{N}$

Using the kinematic equation $v = u + at$:

$$18 \text{m/s} = 0 \text{m/s} + a \times 10 \text{s}$$ $$18 = 10a$$ $$a = \frac{18}{10} \text{m/s}^2 = 1.8 \text{m/s}^2$$

$$F_{net} = 1600 \text{kg} \times 1.8 \text{m/s}^2$$ $$F_{net} = 2880 \text{N}$$

The net force is the applied force minus the frictional force ($F_{net} = F_{applied} - F_f$).

$$F_{applied} = F_{net} + F_f$$ $$F_{applied} = 2880 \text{N} + 500 \text{N}$$ $$F_{applied} = 3380 \text{N}$$

The force required to accelerate the car is $\boxed{3380 \text{N}}$.

Using the equation $s = ut + \frac{1}{2}at^2$:

$$s = (0 \text{m/s})(10 \text{s}) + \frac{1}{2}(1.8 \text{m/s}^2)(10 \text{s})^2$$ $$s = 0 + \frac{1}{2}(1.8)(100)$$ $$s = 0.9 \times 100$$ $$s = 90 \text{m}$$

The distance traveled is $\boxed{90 \text{m}}$.