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

The formula for average speed is $v=\frac{d}{t}$, where $v$ is average speed, $d$ is distance traveled, and $t$ is time elapsed.

Step2: Answer part a

From the graph, at $t = 0$, $d=0$ and at $t = 2$, $d = 80$ km. So the distance traveled in the first two - hours is $d=80 - 0=80$ km.

Step3: Answer part b

Using the average - speed formula $v=\frac{d}{t}$, with $d = 80$ km and $t = 2$ hours, we have $v=\frac{80}{2}=40$ km/h.

Step4: Answer part c

At $t = 2$, $d = 80$ km and at $t = 4$, $d = 100$ km. The distance traveled $d=100 - 80 = 20$ km, and $t=4 - 2=2$ hours. So $v=\frac{20}{2}=10$ km/h.

Step5: Answer part d

At $t = 4$, $d = 100$ km and at $t = 5$, $d = 100$ km. The distance traveled $d=100 - 100 = 0$ km, and $t=5 - 4 = 1$ hour. So $v=\frac{0}{1}=0$ km/h.

Step6: Answer part e

The instantaneous speed at $t = 7$ is the slope of the tangent line to the curve at $t = 7$. Since the graph is a straight - line segment from $t = 5$ to $t = 9$, the slope of this line is constant. The distance at $t = 5$ is $d = 100$ km and at $t = 9$ is $d = 180$ km. The slope $m=\frac{180 - 100}{9 - 5}=\frac{80}{4}=20$ km/h. So the instantaneous speed at $t = 7$ is 20 km/h.

Step7: Answer part f

At $t = 2$, $d = 80$ km and at $t = 5$, $d = 100$ km. The distance traveled $d=100 - 80 = 20$ km, and $t=5 - 2=3$ hours. So $v=\frac{20}{3}\approx6.67$ km/h.

Step8: Answer part g

At $t = 0$, $d = 0$ km and at $t = 9$, $d = 180$ km. Using the average - speed formula $v=\frac{d}{t}$, with $d = 180$ km and $t = 9$ hours, we have $v=\frac{180}{9}=20$ 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