1. **State the problem:** We are given the velocity function of a particle as $v(t) = t^3 - 10t^2 + 29t - 20$ feet per second, and we need to find the displacement of the particle over the time interval $0 \leq t \leq 5$. 2. **Recall the relationship:** Displacement is the integral of velocity over the given time interval. So, displacement $= \int_0^5 v(t) \, dt = \int_0^5 (t^3 - 10t^2 + 29t - 20) \, dt$. 3. **Integrate the velocity function:** $$ \int (t^3 - 10t^2 + 29t - 20) \, dt = \frac{t^4}{4} - \frac{10t^3}{3} + \frac{29t^2}{2} - 20t + C $$ 4. **Evaluate the definite integral from 0 to 5:** $$ \left[ \frac{t^4}{4} - \frac{10t^3}{3} + \frac{29t^2}{2} - 20t \right]_0^5 = \left( \frac{5^4}{4} - \frac{10 \times 5^3}{3} + \frac{29 \times 5^2}{2} - 20 \times 5 \right) - 0 $$ Calculate each term: - $\frac{5^4}{4} = \frac{625}{4} = 156.25$ - $\frac{10 \times 5^3}{3} = \frac{10 \times 125}{3} = \frac{1250}{3} \approx 416.67$ - $\frac{29 \times 5^2}{2} = \frac{29 \times 25}{2} = \frac{725}{2} = 362.5$ - $20 \times 5 = 100$ So, $$ 156.25 - 416.67 + 362.5 - 100 = (156.25 + 362.5) - (416.67 + 100) = 518.75 - 516.67 = 2.08 $$ 5. **Convert to fraction:** $2.08 = \frac{32}{15}$ approximately, but let's check exact fraction: Calculate exactly: $$ \frac{625}{4} - \frac{1250}{3} + \frac{725}{2} - 100 = \frac{625}{4} - \frac{1250}{3} + \frac{725}{2} - \frac{400}{4} $$ Find common denominator 12: $$ \frac{625 \times 3}{12} - \frac{1250 \times 4}{12} + \frac{725 \times 6}{12} - \frac{400 \times 3}{12} = \frac{1875 - 5000 + 4350 - 1200}{12} = \frac{1875 + 4350 - 5000 - 1200}{12} = \frac{6225 - 6200}{12} = \frac{25}{12} $$ 6. **Final answer:** The displacement is $\frac{25}{12}$ feet, which is approximately 2.08 feet. 7. **Check options:** None of the options exactly match $\frac{25}{12}$. Let's verify if the problem expects total displacement or net displacement. Since velocity changes sign, displacement is net area under velocity curve. Since the problem asks for displacement, the integral is correct. The closest option is none, so the answer is $\frac{25}{12}$ feet. **Note:** The options given do not match the exact displacement calculated. **Summary:** The displacement of the particle on $0 \leq t \leq 5$ is $\boxed{\frac{25}{12}}$ feet.