Senior Three Physics Notes

Key ConceptsThis chapter covers how to read displacement-time and velocity-time graphs, and how to use SUVAT equations to solve motion problems. For exams, you must convert what the graph “says” (area/slope) into the correct variables (s, u, v, a, t).

Core ideas you MUST know:

Displacement-time (s-t) graphs

Gradient (slope) = velocity (constant gradient → constant velocity)

Area under the graph = not used for s-t unless you have a v-t graph; for s-t, read displacement directly at times

Velocity-time (v-t) graphs

Gradient (slope) = acceleration

Area under v-t graph = displacement

Constant acceleration patterns

v-t graph is a straight line when acceleration is constant

s-t graph is a curve (parabola) when acceleration is constant (because velocity changes)

SUVAT equations (constant acceleration)

Use only if acceleration is constant and motion is straight-line

Choose the equation that matches the given variables (avoid mixing cases)

Sign convention

Decide positive direction; velocities/displacements opposite direction become negative

Before you continue:If a v-t graph is a horizontal line at +3 m/s for 5 s, what are the acceleration and displacement?

Key TermsDisplacement - change in position (can be negative depending on chosen direction).

Velocity - rate of change of displacement; includes direction (m/s).

Acceleration - rate of change of velocity; can be positive or negative (m/s^2).

Displacement-time graph (s-t) - graph of displacement vs time.

Velocity-time graph (v-t) - graph of velocity vs time.

Gradient (slope) - change in y divided by change in x on a graph.

Area under a v-t graph - equals displacement during that time interval.

Active RecallDisplacement - __________

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Velocity - __________

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Acceleration - __________

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Displacement-time graph (s-t) - __________

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Velocity-time graph (v-t) - __________

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Gradient (slope) - __________

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Area under a v-t graph - __________

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Worked ExamplesExample 1: Read acceleration from a v-t graph (straight line)A car moves with velocity increasing from 2 m/s to 10 m/s in 4 s. Find the acceleration.

Use acceleration = gradient of v-t graph = (final v - initial u) / t

a = (10 - 2) / 4 = 8 / 4 = 2 m/s^2

Acceleration is positive because velocity increases in the positive direction.

Now you tryA runner’s velocity increases from 6 m/s to 18 m/s in 3 s. Find the acceleration.

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Example 2: Displacement from area under a v-t graphA bus slows from 20 m/s to 10 m/s uniformly over 5 s. Find the displacement.

For constant acceleration, v-t graph is a trapezium.

Displacement = area = (1/2) x (sum of parallel sides) x height

Parallel sides are 20 and 10; height is 5 s

s = (1/2) x (20 + 10) x 5 = (1/2) x 30 x 5 = 75 m

Now you tryA bicycle speeds up uniformly from 4 m/s to 14 m/s in 6 s. Find the displacement.

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Example 3: SUVAT equation selectionA train accelerates from rest (u = 0) to 25 m/s in 5 s with constant acceleration. Find acceleration and distance.

Known: u = 0, v = 25, t = 5

a = (v - u) / t = (25 - 0) / 5 = 5 m/s^2

Use s = (u + v)/2 x t for constant acceleration

s = (0 + 25)/2 x 5 = 12.5 x 5 = 62.5 m

Now you tryA scooter starts at 3 m/s and reaches 15 m/s after 4 s (constant acceleration). Find distance travelled.

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Practice Questions(Easy) A displacement-time graph is a straight line increasing. State what this tells you about the acceleration.

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(Easy) A v-t graph is a horizontal line at -2 m/s for 8 s....