Question

1. describe an example (with numerical values) where an object has an average velocity is zero but the average speed is 100 km/h

Explanation:

Step1: Recall definitions

Average velocity is displacement over time, average speed is distance over time. Displacement is zero when object returns to start.

Step2: Construct example

Let’s take a car. It travels 100 km east at 100 km/h (time \( t_1 = \frac{100\ \text{km}}{100\ \text{km/h}} = 1\ \text{h} \)), then 100 km west at 100 km/h (time \( t_2 = \frac{100\ \text{km}}{100\ \text{km/h}} = 1\ \text{h} \)).

Step3: Calculate displacement and distance

Total displacement: \( 100 - 100 = 0\ \text{km} \). Total distance: \( 100 + 100 = 200\ \text{km} \). Total time: \( 1 + 1 = 2\ \text{h} \).

Step4: Calculate velocity and speed

Average velocity: \( \frac{0\ \text{km}}{2\ \text{h}} = 0\ \text{km/h} \). Average speed: \( \frac{200\ \text{km}}{2\ \text{h}} = 100\ \text{km/h} \).

Answer:

Consider a car that travels 100 km east at a speed of 100 km/h (taking 1 hour) and then travels 100 km west at a speed of 100 km/h (taking another 1 hour). The total displacement of the car is \( 100\ \text{km} - 100\ \text{km} = 0\ \text{km} \), so the average velocity (displacement divided by total time) is \( \frac{0\ \text{km}}{1\ \text{h}+1\ \text{h}} = 0\ \text{km/h} \). The total distance traveled is \( 100\ \text{km} + 100\ \text{km} = 200\ \text{km} \), and the total time is \( 2\ \text{h} \), so the average speed (distance divided by total time) is \( \frac{200\ \text{km}}{2\ \text{h}} = 100\ \text{km/h} \).