Describing motion
Section: Kinematics | Syllabus: Cambridge AS Level Physics 9702
Describing Motion To analyze how objects move, we use specific physical quantities. These are categorized into scalars (magnitude only) and vectors (magnitude and direction). Scalar Quantities Vector Quantities Distance (d): The length of the path travelled (m).
Displacement (s): Distance travelled in a particular direction (m). Speed (v): Change in distance per unit time (m s^-1). Velocity (v/u): Change in displacement per unit time (m s^-1). - Acceleration (a): Change in velocity per unit time (m s^-2).
Distance, Displacement, Speed, and Velocity For an object travelling in a straight line at a constant velocity, the displacement s is the product of velocity v and time t . s = v × t (Applies only to constant velocity) Direction Matters In a straight road, distance and displacement may be the same.
However, if an object returns to its start point, its total distance is the path length, but its displacement is zero . Average Speed and Average Velocity If velocity is changing, we use the average value over a time interval Δ t (where Δ means "change in").
Average Velocity (v_av) = fraction (Change in displacement / Time) Average Speed = Total DistanceTotal Time Worked Example: Unit Conversions Question: A motorcyclist starts from rest and travels 20 km along a straight road in 30 minutes.
Calculate the average speed in: (i) km h^-1 (ii) m s^-1. Solution (i) In km h^-1: Time = 30 mins = 0.5 hours. v = 20 km0.5 h = 40 km h^-1 (ii) In m s^-1: Distance = 20,000 m; Time = 30 × 60 = 1800 s. v = 20000 m1800 s = 11.1 m s^-1 Note: The speed must have been greater than the average at some points, as the cyclist started from rest (0 m s^-1).
Worked Example: Thinking Distance Question: A car travels at 72 km h^-1. Calculate the distance it travels in a thinking time of 0.25 s. Solution Step 1: Convert speed to m s^-1 72 km = 72,000 m; 1 hour = 3600 s.
v = fraction = 20 m s^-1 Step 2: Calculate distance d = v × t = 20 × 0.25 = 5.0 m Displacement-Time (s-t) Graphs The gradient of a displacement-time graph represents velocity . Straight line: Constant velocity.
Steeper slope: Higher velocity. Horizontal line: Zero velocity (at rest). Curve: Changing velocity (acceleration). [Diagram: Displacement-Time Graph] (Straight line through origin for constant velocity; Curve upwards for acceleration) Instantaneous Velocity Using Tangents When an object accelerates, its s-t graph is curved.
To find the velocity at a specific instant : Draw a tangent (a straight line just touching the curve) at the desired time. Construct a large gradient triangle. Calculate the gradient (fraction) of that tangent.
Worked Example: Finding Velocity from Curve Question: An aeroplane on a runway has a curved s-t graph. At t = 20 s, a tangent is drawn. The tangent passes through (10 s, 800 m) and (40 s, 2000 m). Calculate the velocity.
Solution Δ s = 2000 - 800 = 1200 m Δ t = 40 - 10 = 30 s v = fraction = 40 m s^-1 Positive and Negative Displacement As displacement is a vector, we assign signs to directions: Horizontal: Right (+) and Left (-).
Vertical: Up (+) and Down (-). A ball thrown upwards has positive displacement until it falls below its starting point, where its displacement becomes negative . [Diagram: Multi-directional s-t Graph] (Graph crossing the time axis into negative values as an object falls below the release point) Velocity-Time (v-t) Graphs A velocity-time graph provides two critical pieces of information: Gradient = Acceleration Area under the graph = Displacement (s) Uniform vs Non-Uniform Acceleration A straight line on a v-t graph indicates uniform (constant) acceleration .
A curve indicates non-uniform acceleration . Area Under Velocity-Time Graphs For constant velocity : Area is a rectangle (s = v × t). For uniform acceleration from rest : Area is a triangle (s = fraction × base × height).
For non-uniform acceleration : Count the squares on the graph paper. Multiply the number of squares by the displacement represented by a single square. Worked Example: Interpreting v-t Graph Question: A cyclist accelerates from 0 to 10 m s^-1 in 10 s, travels at constant velocity for 15 s, and then decelerates to a stop in 8 s.
Calculate: (i) Total displacement. (ii) Acceleration in the last 8 s. Solution (i) Total Displacement (Area): Part 1 (Triangle): fraction × 10 × 10 = 50 m Part 2 (Rectangle): 15 × 10 = 150 m Part 3 (Triangle): fraction × 8 × 10 = 40 m Total s = 50 + 150 + 40 = 240 m (ii) Acceleration in last 8 s (Gradient): a = fraction = fraction = -1.25 m s^-2 (Negative sign indicates deceleration) Key Exam Takeaways Definitions: Displacement and Velocity are vectors; Distance and Speed are scalars.
Graph 1: Gradient of s-t graph = Velocity. Graph 2: Gradient of v-t graph = Acceleration. Graph 3: Area under v-t graph = Displacement. Complex Motion: For curves, use tangents for velocity (s-t) or squares for displacement (v-t).
Rest: "At rest" means velocity = 0.
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