Step-by-step solutions to problems on velocity and speed of moving objects. More tutorials are available.
A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction.
a) What is the man's average speed for the whole journey?
b) What is the man's average velocity for the whole journey?
Solution to Problem 1:
a) Average speed:
\[ \text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{7\,\text{km} + 2\,\text{km}}{2\,\text{h} + 1\,\text{h}} = \frac{9\,\text{km}}{3\,\text{h}} = 3\,\text{km/h} \]
b) Average velocity (same direction, so displacement = total distance):
\[ \text{average velocity} = \frac{\text{displacement}}{\text{total time}} = \frac{9\,\text{km}}{3\,\text{h}} = 3\,\text{km/h} \]
A man walks 7 km East in 2 hours and then 2.5 km West in 1 hour.
a) What is the man's average speed for the whole journey?
b) What is the man's average velocity for the whole journey?
Solution to Problem 2:
a) Average speed:
\[ \text{average speed} = \frac{7\,\text{km} + 2.5\,\text{km}}{2\,\text{h} + 1\,\text{h}} = \frac{9.5\,\text{km}}{3\,\text{h}} \approx 3.2\,\text{km/h} \]
b) Displacement = \(7\,\text{km (East)} - 2.5\,\text{km (West)} = 4.5\,\text{km (East)}\):
\[ \text{average velocity} = \frac{4.5\,\text{km}}{3\,\text{h}} = 1.5\,\text{km/h (East)} \]
You start walking from a point on a circular field of radius 0.5 km and 1 hour later you are at the same point.
a) What is your average speed?
b) What is your average velocity?
Solution to Problem 3:
a) Distance = circumference = \(2\pi r = 2\pi(0.5\,\text{km}) = \pi\,\text{km}\):
\[ \text{average speed} = \frac{\pi\,\text{km}}{1\,\text{h}} \approx 3.14\,\text{km/h} \]
b) Displacement = 0 (same starting and ending point):
\[ \text{average velocity} = 0 \]
John drove South 120 km at 60 km/h and then East 150 km at 50 km/h.
a) What is the average speed for the whole journey?
b) What is the magnitude of the average velocity?
Solution to Problem 4:
a) Total time = \(\frac{120}{60} + \frac{150}{50} = 2 + 3 = 5\,\text{h}\). Total distance = \(120 + 150 = 270\,\text{km}\):
\[ \text{average speed} = \frac{270\,\text{km}}{5\,\text{h}} = 54\,\text{km/h} \]
b) Displacement magnitude (Pythagorean theorem):
\[ \text{displacement} = \sqrt{120^2 + 150^2} = \sqrt{14400 + 22500} = \sqrt{36900} = 30\sqrt{41}\,\text{km} \]
\[ \text{average velocity} = \frac{30\sqrt{41}\,\text{km}}{5\,\text{h}} \approx 38.4\,\text{km/h} \]
If I can walk at an average speed of 5 km/h, how many miles can I walk in two hours?
Solution to Problem 5:
\[ \text{distance} = 5\,\text{km/h} \times 2\,\text{h} = 10\,\text{km} \]
Convert to miles (1 km ≈ 0.62 miles):
\[ 10\,\text{km} \times 0.62\,\text{miles/km} = 6.2\,\text{miles} \]
A train travels a distance \(d\) at 60 mi/h, then another \(2d\) at 80 mi/h in the same direction.
What is the average speed for the whole journey?
Solution to Problem 6:
Total time = \(\frac{d}{60} + \frac{2d}{80} = \frac{d}{60} + \frac{d}{40} = d\left(\frac{1}{60}+\frac{1}{40}\right) = d\left(\frac{2+3}{120}\right) = \frac{5d}{120} = \frac{d}{24}\,\text{h}\):
Total distance = \(d + 2d = 3d\):
\[ \text{average speed} = \frac{3d}{d/24} = 3d \times \frac{24}{d} = 72\,\text{mi/h} \]
A car travels 22 km south, 12 km west, and 14 km north in half an hour.
a) What is the average speed?
b) What is the final displacement?
c) What is the average velocity?
Solution to Problem 7:
a) Total distance = \(22 + 12 + 14 = 48\,\text{km}\):
\[ \text{average speed} = \frac{48\,\text{km}}{0.5\,\text{h}} = 96\,\text{km/h} \]
b) Net south displacement = \(22 - 14 = 8\,\text{km}\). West displacement = \(12\,\text{km}\).
Displacement magnitude = \(\sqrt{8^2 + 12^2} = \sqrt{64 + 144} = \sqrt{208} = 4\sqrt{13}\,\text{km}\).
c) Average velocity magnitude:
\[ \text{average velocity} = \frac{4\sqrt{13}\,\text{km}}{0.5\,\text{h}} \approx 28.8\,\text{km/h} \]
A man walks from point A to F along a grid in 3250 seconds.
a) What is the average speed in m/s?
b) What is the displacement magnitude?
c) What is the average velocity magnitude in m/s?
Solution to Problem 8:
a) Total distance = AB + BC + CD + DE + EF = \(3 + 1 + 1.5 + 0.5 + 0.5 = 6.5\,\text{km} = 6500\,\text{m}\):
\[ \text{average speed} = \frac{6500\,\text{m}}{3250\,\text{s}} = 2\,\text{m/s} \]
b) Displacement (AF) magnitude (Pythagorean theorem):
\[ \text{displacement} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,\text{km} = 5000\,\text{m} \]
c) Average velocity magnitude:
\[ \text{average velocity} = \frac{5000\,\text{m}}{3250\,\text{s}} \approx 1.54\,\text{m/s} \]