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Motion is the change in position of an object with respect to time. When something moves, it changes its place or location. Understanding motion is important because it helps us describe and predict how objects behave in the world around us-from a car driving on the road to a ball thrown in the air.
To study motion, we use several key concepts:
We will explore each of these concepts carefully, using examples and diagrams to make them clear.
Distance is the total length of the path traveled by an object, no matter which direction it moves. It is a scalar quantity, which means it only has magnitude (size) but no direction.
Displacement is the shortest straight-line distance from the starting point to the ending point of the motion, along with the direction. It is a vector quantity, meaning it has both magnitude and direction.
For example, if you walk 3 meters east and then 4 meters north, your total distance traveled is 7 meters (3 + 4), but your displacement is the straight-line distance from your starting point to your final position, which can be found using the Pythagorean theorem.
In this diagram, the object moves from the start point up 60 meters, then right 100 meters. The total distance traveled is 160 meters (60 + 100), but the displacement is the straight line from start to end.
Speed tells us how fast an object is moving. It is the rate at which distance is covered. Since speed does not include direction, it is a scalar quantity.
Velocity is speed with a direction. It tells us how fast and in which direction an object is moving. Velocity is a vector quantity.
Both speed and velocity can be average or instantaneous:
| Property | Speed | Velocity |
|---|---|---|
| Definition | Rate of change of distance | Rate of change of displacement |
| Formula (average) | \( v = \frac{d}{t} \) | \( \vec{v} = \frac{\vec{s}}{t} \) |
| Quantity type | Scalar (magnitude only) | Vector (magnitude and direction) |
| Can be negative? | No | Yes, depending on direction |
| Example | Car moving at 60 km/h | Car moving 60 km/h east |
Acceleration is the rate at which velocity changes with time. It tells us how quickly an object speeds up, slows down, or changes direction.
Acceleration is a vector quantity because velocity is a vector. When acceleration and velocity are in the same direction, the object speeds up (positive acceleration). When acceleration is opposite to velocity, the object slows down (negative acceleration or deceleration).
In this velocity-time graph, velocity increases steadily over time, showing constant acceleration.
Step 1: Calculate total distance.
Distance = 5 m + 12 m = 17 m
Step 2: Calculate displacement using Pythagoras theorem.
Displacement \( s = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \) meters
Step 3: Direction of displacement is northeast (45° between east and north).
Answer: Distance = 17 m, Displacement = 13 m northeast.
Step 1: Calculate total distance.
Total distance = 10 km + 10 km = 20 km
Step 2: Calculate total time.
Total time = 30 min + 20 min = 50 min = \(\frac{50}{60}\) hours = \(\frac{5}{6}\) hours
Step 3: Calculate average speed.
\( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{20}{5/6} = 20 \times \frac{6}{5} = 24 \text{ km/h} \)
Step 4: Calculate displacement.
Since the cyclist returns to the starting point, displacement = 0 km.
Step 5: Calculate average velocity.
\( \text{Average velocity} = \frac{\text{Displacement}}{\text{Total time}} = \frac{0}{5/6} = 0 \text{ km/h} \)
Answer: Average speed = 24 km/h, Average velocity = 0 km/h.
Step 1: Identify initial velocity \( v_i = 20 \, m/s \), final velocity \( v_f = 40 \, m/s \), and time \( t = 5 \, s \).
Step 2: Use acceleration formula:
\[ a = \frac{v_f - v_i}{t} = \frac{40 - 20}{5} = \frac{20}{5} = 4 \, m/s^2 \]
Answer: Acceleration = \(4 \, m/s^2\).
Step 1: Calculate acceleration using velocity change and time.
\[ a = \frac{v_f - v_i}{t} = \frac{20 - 0}{10} = 2 \, m/s^2 \]
Step 2: Calculate displacement from velocity-time graph area (triangle area):
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 20 = 100 \, m\)
Answer: Acceleration = \(2 \, m/s^2\), Displacement = 100 m.
Step 1: Understand that acceleration is the rate of change of velocity with respect to time, i.e., the derivative of velocity.
Step 2: Differentiate \( v = 3t^2 + 2 \) with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 + 2) = 6t \]
Step 3: Substitute \( t = 4 \) seconds:
\[ a = 6 \times 4 = 24 \, m/s^2 \]
Answer: Acceleration at 4 seconds is \(24 \, m/s^2\).
When to use: When distinguishing between distance and displacement in problems.
When to use: When solving graph-based motion problems.
When to use: Always before solving numerical problems.
When to use: When dealing with round trip motion problems.
When to use: When calculating acceleration from velocity data.
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