Concept:Average velocity depends on the net displacement vector rather than the total distance traveled.
Formula:$$\vec{v}_{\text{avg}} = \frac{\vec{d}_{\text{net}}}{t}$$
Solution:Let's define East as the positive x-direction and West as the negative x-direction. Each trip occurs over a duration of \(t = 1\text{ hour}\).
Let's calculate the net displacement for the journeys described in
Option B:
- Car 1: Travels +40 km, then -20 km:
$$\vec{d}_{\text{net1}} = 40\text{ km} - 20\text{ km} = +20\text{ km (East)}$$
$$\vec{v}_{\text{avg1}} = \frac{20\text{ km}}{1\text{ hr}} = 20\text{ km/hr East}$$ - Car 2: Travels +20 km:
$$\vec{d}_{\text{net2}} = +20\text{ km (East)}$$
$$\vec{v}_{\text{avg2}} = \frac{20\text{ km}}{1\text{ hr}} = 20\text{ km/hr East}$$
Both cars have the exact same net displacement vector over the same time interval, so they have identical average velocities.
Why other options are incorrect:- A, C, and D contain pairs of journeys with unequal net displacements (for example, in D, Car 1 has a net displacement of 0 km, whereas Car 2 has a net displacement of 20 km).
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