Question

use the graph below to answer the following questions: (express all answers in kilometers and hours.) train trip a. how far did the train travel during the first two hours? b. what was the average speed during the first two hours? c. what was the average speed of the train between the second and the fourth hour? d. what was the average speed of the train between hour 4 and hour 5? e. what was the instantaneous speed of the train at hour 7? f. what was the average speed of the train between the second and the fifth hour? g. what is the average speed of the train for the entire trip?

Explanation:

Step1: Recall speed - distance formula

Average speed $v=\frac{d}{t}$, where $d$ is distance and $t$ is time.

Step2: Answer part a

From the graph, at $t = 2$ hours, $d=80$ km. So the train traveled 80 km in the first two - hours.

Step3: Answer part b

Using $v=\frac{d}{t}$, with $d = 80$ km and $t = 2$ hours, $v=\frac{80}{2}=40$ km/h.

Step4: Answer part c

At $t = 2$ hours, $d_1 = 80$ km; at $t = 4$ hours, $d_2 = 100$ km. $\Delta d=d_2 - d_1=100 - 80 = 20$ km, $\Delta t=4 - 2 = 2$ hours. Then $v=\frac{\Delta d}{\Delta t}=\frac{20}{2}=10$ km/h.

Step5: Answer part d

At $t = 4$ hours, $d_1 = 100$ km; at $t = 5$ hours, $d_2 = 100$ km. $\Delta d=d_2 - d_1=100 - 100 = 0$ km, $\Delta t=5 - 4 = 1$ hour. So $v=\frac{\Delta d}{\Delta t}=0$ km/h.

Step6: Answer part e

The instantaneous speed at $t = 7$ hours is the slope of the tangent line at $t = 7$ on the distance - time graph. The graph is a straight - line from $t = 5$ to $t = 9$ hours. The slope of this line: at $t = 5$ hours, $d_1 = 100$ km; at $t = 9$ hours, $d_2 = 180$ km. $\Delta d=d_2 - d_1=180 - 100 = 80$ km, $\Delta t=9 - 5 = 4$ hours. The slope (instantaneous speed at any point on this straight - line segment) is $v=\frac{\Delta d}{\Delta t}=\frac{80}{4}=20$ km/h.

Step7: Answer part f

At $t = 2$ hours, $d_1 = 80$ km; at $t = 5$ hours, $d_2 = 100$ km. $\Delta d=d_2 - d_1=100 - 80 = 20$ km, $\Delta t=5 - 2 = 3$ hours. Then $v=\frac{\Delta d}{\Delta t}=\frac{20}{3}\approx6.67$ km/h.

Step8: Answer part g

At $t = 0$ hours, $d_1 = 0$ km; at $t = 9$ hours, $d_2 = 180$ km. $\Delta d=d_2 - d_1=180$ km, $\Delta t=9$ hours. Then $v=\frac{\Delta d}{\Delta t}=\frac{180}{9}=20$ km/h.

Answer:

a. 80 km
b. 40 km/h
c. 10 km/h
d. 0 km/h
e. 20 km/h
f. $\frac{20}{3}\approx6.67$ km/h
g. 20 km/h