1. **Problem statement:** We analyze a cyclist's journey from home to a city and back using the given distance-time graph. 2. **Given data from the graph:** - From 0 to 1 hour: distance increases from 0 to 15 miles. - From 1 to 2 hours: distance remains at 15 miles (stationary). - From 2 to 3 hours: distance increases from 15 to 35 miles. - From 3 to 4 hours: distance remains at 35 miles (stationary). - From 4 to 6 hours: distance decreases from 35 to 0 miles (return trip). 3. **(a) Total stationary time:** Stationary means distance does not change over time. - From 1 to 2 hours: stationary for $2 - 1 = 1$ hour. - From 3 to 4 hours: stationary for $4 - 3 = 1$ hour. Total stationary time = $1 + 1 = 2$ hours. 4. **(b) Total distance traveled:** Distance traveled is the sum of all distances covered, regardless of direction. - From 0 to 1 hour: $15 - 0 = 15$ miles. - From 1 to 2 hours: $0$ miles (stationary). - From 2 to 3 hours: $35 - 15 = 20$ miles. - From 3 to 4 hours: $0$ miles (stationary). - From 4 to 6 hours: $35 - 0 = 35$ miles (return trip). Total distance = $15 + 0 + 20 + 0 + 35 = 70$ miles. 5. **(c) Speed on the way home:** The way home is from 4 to 6 hours, distance decreases from 35 to 0 miles. Speed = distance / time = $\frac{35 - 0}{6 - 4} = \frac{35}{2} = 17.5$ miles per hour. 6. **(d) Greatest speed:** Speed is the slope of the distance-time graph. Calculate speeds for each moving segment: - 0 to 1 hour: $\frac{15 - 0}{1 - 0} = 15$ mph. - 2 to 3 hours: $\frac{35 - 15}{3 - 2} = 20$ mph. - 4 to 6 hours: $\frac{0 - 35}{6 - 4} = -17.5$ mph (speed magnitude 17.5 mph). Greatest speed is $20$ mph (between 2 and 3 hours). **Final answers:** - (a) $2$ hours stationary. - (b) $70$ miles traveled. - (c) $17.5$ mph on the way home. - (d) Greatest speed $20$ mph.