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Example · Example 2

Q.A car travels 400 m due east along a straight road in 20 s, then reverses direction and travels 150 m due west in 10 s. Find

(a) the average speed and
(b) the average velocity of the car for the entire 30 s journey.
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Setting up. Take east as the positive direction. The car travels 400 m east in 20 s, then reverses and travels 150 m west in the next 10 s. Total time Δt=20+10=30\Delta t = 20 + 10 = 30 s.

Average speed. This uses the total path length covered, regardless of direction:

total path length=400 m+150 m=550 m\text{total path length} = 400\ \text{m} + 150\ \text{m} = 550\ \text{m}

average speed=550 m30 s≈18.3 m/s\text{average speed} = \frac{550\ \text{m}}{30\ \text{s}} \approx 18.3\ \text{m/s}

Average velocity. This uses the net displacement — the car's final position relative to where it started. Taking east as positive, the car moves +400+400 m and then −150-150 m, so its net displacement is

Δx=400−150=250 m (east)\Delta x = 400 - 150 = 250\ \text{m (east)}

vˉ=ΔxΔt=250 m30 s≈8.3 m/s, directed east\bar v = \frac{\Delta x}{\Delta t} = \frac{250\ \text{m}}{30\ \text{s}} \approx 8.3\ \text{m/s, directed east}

Comparison. As expected from Section 2.3, the average speed (≈18.3\approx 18.3 m/s) is larger than the magnitude of the average velocity (≈8.3\approx 8.3 m/s), because the car reversed direction partway through its journey — the westward leg adds fully to the path length but partially cancels the eastward displacement.

[!ANSWER] Average speed ≈ 18.3 m/s; average velocity ≈ 8.3 m/s, directed east.

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