Question

this graph shows a trip on a bike trail. the trail has markers every 0.5 km showing the distance from the beginning of the trail.
column a

1. 1. when was the bike rider going the fastest?
2. 2. when was the bike rider going the slowest?
3. 3. during what time was the longest the rider stopped for?
4. 4. during what time was the shortest the rider stopped for?
5. 5. during what time was the rider going away from the beginning of the trail fastest?
6. 6. during what time was the rider going back towards the beginning of the trail the slowest?

column b
a. 2.4 - 2.6 hours
b. 0 - 0.8 hours
c. 2.2 - 2.4 hours
d. 1.4 - 2.2 hours
e. 0.8 - 1.4 hours
f. 2.6 - 3 hours

Explanation:

Step1: Analyze speed from slope

Speed = $\frac{\text{Change in Distance}}{\text{Change in Time}}$, steeper slope = faster speed.

Step2: Match Q1 (fastest overall)

The steepest slope is 2.4-2.6 hours (distance drops from 9.5 to 5 km in 0.2h, speed = $\frac{9.5-5}{2.6-2.4}=22.5$ km/h). Match to a.

Step3: Match Q2 (slowest moving)

The shallowest positive/negative slope is 1.4-2.2 hours (distance rises from 8 to 9.5 km in 0.8h, speed = $\frac{9.5-8}{2.2-1.4}=1.875$ km/h). Match to d.

Step4: Match Q3 (longest stop)

Flat line (no distance change) 0.8-1.4 hours, duration = $1.4-0.8=0.6$ hours. Match to e.

Step5: Match Q4 (shortest stop)

Flat line 2.2-2.4 hours, duration = $2.4-2.2=0.2$ hours. Match to c.

Step6: Match Q5 (fastest away)

Steepest positive slope 0-0.8 hours (distance rises to 8 km in 0.8h, speed = $\frac{8-0}{0.8-0}=10$ km/h). Match to b.

Step7: Match Q6 (slowest back)

Shallowest negative slope 2.6-3 hours (distance drops from 5 to 0 km in 0.4h, speed = $\frac{5-0}{3-2.6}=12.5$ km/h, slower than 2.4-2.6h). Match to f.

Answer:

1. a. 2.4 - 2.6 hours
2. d. 1.4 - 2.2 hours
3. e. 0.8 - 1.4 hours
4. c. 2.2 - 2.4 hours
5. b. 0 - 0.8 hours
6. f. 2.6 - 3 hours