Let's solve the problem step by step for both parts (a) and (b): --- ## Part (a): Distance traveled during the first 30 minutes ### Step 1: Set up the integral for distance The distance traveled is the integral of velocity \( v(t) \) from \( t= \) to \( t=30 \): \[ \text{Distance} = \int_{}^{30} v(t)\,dt = \int_{}^{30} \frac{1}{2}t\,dt \] **Explanation:** We integrate the velocity function over the time interval of interest ( to 30 minutes). --- ### Step 2: Compute the integral \[ \int_{}^{30} \frac{1}{2}t\,dt = \frac{1}{2} \int_{}^{30} t\,dt = \frac{1}{2} \left[\frac{t^2}{2}\right]_{}^{30} = \frac{1}{2} \left(\frac{30^2}{2} - \frac{^2}{2}\right) = \frac{1}{2} \times \frac{900}{2} = \frac{900}{4} = 225 \] So, the distance traveled in the first 30 minutes is **225 feet**. --- ## Part (b): Distance traveled between the first and second hour ### Step 1: Set up the integral for the time interval 1 hour = 60 minutes. We want the distance from \( t=60 \) to \( t=120 \): \[ \text{Distance} = \int_{60}^{120} v(t)\,dt = \int_{60}^{120} \frac{1}{2}t\,dt \] **Explanation:** We are calculating the area under the velocity curve between 60 and 120 minutes. --- ### Step 2: Compute the integral \[ \int_{60}^{120} \frac{1}{2}t\,dt = \frac{1}{2} \left[\frac{t^2}{2}\right]_{60}^{120} = \frac{1}{2} \left(\frac{120^2}{2} - \frac{60^2}{2}\right) = \frac{1}{2} \left(\frac{14400}{2} - \frac{360}{2}\right) = \frac{1}{2} (720 - 180) = \frac{1}{2} \times 540 = 270 \] So, the distance traveled between the first and second hour is **270 feet**. --- ## **Final Answers:** - **a)** The object travels **225 feet** during the first 30 minutes. - **b)** The object travels **270 feet** between the first hour and the second hour.