Question

gpb.org/physics-motion
work each of the following problems. show all work.

1. a ladybug walks 10 cm forward then 5 cm backward in 20 seconds. what is the average speed of the ladybug?

using the information from the previous question, what is the average velocity of the ladybug?

1. a city bus travels 6 blocks east and 8 blocks north. each block is 100 m long. if the bus travels this distance in 15 minutes, what is the average speed of the bus (m/s)?
2. using the information from the previous question, what is the average velocity of the city bus?

questions continued on next page
dm.03_practice problems

Explanation:

Problem 5 (Average Speed of Ladybug)

Step1: Calculate total distance

The ladybug moves 10 cm forward and 5 cm backward. Total distance is the sum of these distances: \( 10 + 5 = 15 \, \text{cm} \).

Step2: Use speed formula (\( v = \frac{d}{t} \))

Time \( t = 20 \, \text{s} \). So, average speed \( v = \frac{15}{20} = 0.75 \, \text{cm/s} \).

Step1: Calculate displacement

Displacement is final position - initial position. Forward 10 cm, backward 5 cm: \( 10 - 5 = 5 \, \text{cm} \) (forward).

Step2: Use velocity formula (\( v = \frac{\Delta x}{t} \))

Time \( t = 20 \, \text{s} \). So, average velocity \( v = \frac{5}{20} = 0.25 \, \text{cm/s} \) (forward direction).

Step1: Calculate total distance

6 blocks east + 8 blocks north. Each block is 100 m: Total distance \( d = (6 + 8) \times 100 = 1400 \, \text{m} \).

Step2: Convert time to seconds

15 minutes = \( 15 \times 60 = 900 \, \text{s} \).

Step3: Use speed formula (\( v = \frac{d}{t} \))

\( v = \frac{1400}{900} \approx 1.56 \, \text{m/s} \) (rounded to two decimal places).

Answer:

\( 0.75 \, \text{cm/s} \)

Problem 5 (Average Velocity of Ladybug)