Question

g. a cyclist traveled 45 kilometers in 3 hours. what was the cyclists average speed in kilometers per hour? h. the table below shows the distance traveled by a car during three different intervals in a race. calculate the average speed for each interval. trial 1 2 3 time (hr) 0.5 1.25 1.5 distance (km) 30 45 30 i. a student recorded the time it took a ball to roll a certain distance. the data is represented in the graph below. use the graph to calculate the average speed of the ball. motion of ball distance (m) 0 5 10 15 20 25 0 5 10 15 20 25 time (s) j. during a school science experiment, a group of students timed how long it took a toy car to travel 5 meters on different surfaces. on a smooth surface, it took 2 seconds, and on a rough surface, it took 4 seconds. calculate the average speed for each surface and compare the results. k. a student graphs the motion of a toy car. the student graphs the data from their investigation. which two statements describe the graph? circle your answer. motion of toy car distance (m) 0 5 10 15 20 25 0 10 20 30 40 50 time (s) a. the average speed of the toy car in this investigation is 3.33 m/s. b. the average speed of the toy car in this investigation is 0.3 m/s. c. the toy car decreases speed throughout the investigation. d. the toy car remains at a constant speed throughout the investigation. e. the toy car increases speed throughout the investigation. l. what is the average speed of the toy car?

Explanation:

Step1: Recall average - speed formula

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

Step2: Solve for G

Given $d = 45$ kilometers and $t=3$ hours. Using the formula $v=\frac{d}{t}$, we substitute the values: $v=\frac{45}{3}=15$ kilometers per hour.

Step3: Solve for H

For trial 1: $d = 30$ km and $t = 0.5$ hr. Then $v_1=\frac{30}{0.5}=60$ km/hr.
For trial 2: $d = 45$ km and $t = 1.25$ hr. Then $v_2=\frac{45}{1.25}=36$ km/hr.
For trial 3: $d = 30$ km and $t = 1.5$ hr. Then $v_3=\frac{30}{1.5}=20$ km/hr.

Step4: Solve for I

From the graph, when the ball starts ($t = 0$), $d = 25$ m and when $t = 25$ s, $d = 0$ m. The change in distance $\Delta d=25 - 0=25$ m and the change in time $\Delta t=25 - 0 = 25$ s. Using the average - speed formula $v=\frac{\Delta d}{\Delta t}$, we get $v=\frac{25}{25}=1$ m/s.

Step5: Solve for J

On the smooth surface: $d = 5$ m and $t = 2$ s. Using the formula $v_1=\frac{d}{t}$, we have $v_1=\frac{5}{2}=2.5$ m/s.
On the rough surface: $d = 5$ m and $t = 4$ s. Using the formula $v_2=\frac{d}{t}$, we have $v_2=\frac{5}{4}=1.25$ m/s.

Step6: Solve for K

The average speed of the toy - car in the first graph: When $t = 0$, $d = 25$ m and when $t = 50$ s, $d = 0$ m. $\Delta d=25$ m and $\Delta t = 50$ s. So $v=\frac{25}{50}=0.5$ m/s.
The average speed of the toy - car in the second graph: When $t = 0$, $d = 0$ m and when $t = 15$ s, $d = 10$ m. $\Delta d = 10$ m and $\Delta t=15$ s. So $v=\frac{10}{15}=\frac{2}{3}\approx0.67$ m/s.
The two statements could be:

1. The average speed of the toy - car in the first graph is less than the average speed of the toy - car in the second graph.
2. The distance traveled by the toy - car in the first graph is greater than the distance traveled by the toy - car in the second graph.

Step7: Solve for L

From the graph of the toy - car, when $t = 0$, $d = 25$ m and when $t = 50$ s, $d = 0$ m. Using the average - speed formula $v=\frac{\Delta d}{\Delta t}$, where $\Delta d=25$ m and $\Delta t = 50$ s, we get $v=\frac{25}{50}=0.5$ m/s.

Answer:

G. 15 kilometers per hour
H. Trial 1: 60 km/hr, Trial 2: 36 km/hr, Trial 3: 20 km/hr
I. 1 m/s
J. Smooth surface: 2.5 m/s, Rough surface: 1.25 m/s
K. Two statements: 1. The average speed of the toy - car in the first graph is less than the average speed of the toy - car in the second graph. 2. The distance traveled by the toy - car in the first graph is greater than the distance traveled by the toy - car in the second graph.
L. 0.5 m/s