A car's velocity is given by the function \( v(t) = 5t^2 + 3 \). Calculate the distance traveled by the car from time \( t = 0 \) to \( t = 4 \).

To calculate the distance traveled by the car from time \( t = 0 \) to \( t = 4 \) seconds, we'll integrate the velocity function over this interval. Given the velocity function: \[ v(t) = 5t^2 + 3 \] **Step 1: Set Up the Integral** The distance \( d \) traveled is the integral of velocity with respect to time: \[ d = \int_{0}^{4} v(t) \, dt = \int_{0}^{4} (5t^2 + 3) \, dt \] **Step 2: Integrate the Function** We'll integrate each term separately. 1. **Integrate \( 5t^2 \):** \[ \int 5t^2 \, dt = 5 \cdot \frac{t^3}{3} = \frac{5}{3} t^3 \] 2. **Integrate \( 3 \):** \[ \int 3 \, dt = 3t \] So, the integrated function is: \[ \frac{5}{3} t^3 + 3t \] **Step 3: Evaluate the Integral from 0 to 4** Plug in the upper and lower limits of the integral. 1. **At \( t = 4 \):** \[ \frac{5}{3} (4)^3 + 3(4) = \frac{5}{3} \times 64 + 12 = \frac{320}{3} + 12 = \frac{320}{3} + \frac{36}{3} = \frac{356}{3} \] 2. **At \( t = 0 \):** \[ \frac{5}{3} (0)^3 + 3(0) = 0 + 0 = 0 \] **Step 4: Calculate the Total Distance** Subtract the lower limit result from the upper limit result: \[ d = \frac{356}{3} - 0 = \frac{356}{3} \text{ units} \] **Step 5: Convert to Decimal (Optional)** \[ \frac{356}{3} \approx 118.67 \text{ units} \] **Final Answer:** The car travels a distance of \(\frac{356}{3}\) units, which is approximately 118.67 units, from \( t = 0 \) to \( t = 4 \).