Kinematics: Constant Acceleration and SUVAT

This lesson covers straight-line kinematics with constant acceleration as required by the Edexcel 9MA0 A-Level Mathematics specification. You need to understand displacement-time and velocity-time graphs, and be fluent with the five SUVAT equations.

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Key Quantities

Symbol	Quantity	SI Unit
s	Displacement	metres (m)
u	Initial velocity	m s⁻¹
v	Final velocity	m s⁻¹
a	Acceleration	m s⁻²
t	Time	seconds (s)

Important distinctions:

• Distance is a scalar (always positive); displacement is a vector (can be negative — direction matters).
• Speed is a scalar; velocity is a vector.
• Acceleration is the rate of change of velocity. Negative acceleration (deceleration) means the object is slowing down in the positive direction.

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The SUVAT Equations

These five equations apply only when acceleration is constant (uniform).

1. v = u + at
2. s = ut + ½at²
3. s = vt - ½at²
4. v² = u² + 2as
5. s = ½(u + v)t

Each equation links four of the five SUVAT variables. To solve a problem:

1. List the known quantities and the quantity you need to find.
2. Choose the equation that contains exactly those four variables.

Exam Tip: Always define the positive direction clearly at the start of a problem. This is especially important for vertical motion where you must decide whether upwards or downwards is positive.

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Displacement-Time Graphs

On a displacement-time (s-t) graph:

• The gradient at any point equals the velocity at that instant.
• A straight line means constant velocity (zero acceleration).
• A curve means the velocity is changing (non-zero acceleration).
• A horizontal line means the object is stationary.

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Velocity-Time Graphs

On a velocity-time (v-t) graph:

• The gradient at any point equals the acceleration.
• A horizontal line means constant velocity (zero acceleration).
• A straight line with positive gradient means constant positive acceleration.
• The area under the graph between two times gives the displacement during that interval.

For a trapezium under the graph: Area = ½(u + v) x t

This is consistent with SUVAT equation 5.

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Vertical Motion Under Gravity

Objects falling freely near the Earth's surface have constant acceleration due to gravity:

a = g ≈ 9.8 m s⁻² (downwards)

Convention: If upwards is taken as positive, then a = -9.8 m s⁻².

Example: A ball is thrown vertically upwards at 14.7 m s⁻¹. Taking upwards as positive and g = 9.8 m s⁻²:

To find the time to reach the highest point (v = 0): v = u + at 0 = 14.7 + (-9.8)t t = 14.7/9.8 = 1.5 s

To find the maximum height: s = ut + ½at² s = 14.7(1.5) + ½(-9.8)(1.5²) s = 22.05 - 11.025 = 11.025 m

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Multi-Stage Problems

Some problems involve two or more stages with different accelerations. For each stage:

1. Apply SUVAT separately.
2. The final velocity of one stage becomes the initial velocity of the next.
3. Keep track of the total displacement and total time.

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Deriving the SUVAT Equations

From the definitions:

• Acceleration: a = (v - u)/t, giving v = u + at.
• Displacement is the area under the v-t graph. For constant acceleration, the v-t graph is a straight line from u to v over time t: s = ½(u + v)t
• Substituting v = u + at into s = ½(u + v)t gives s = ut + ½at².
• Eliminating t from v = u + at and s = ½(u + v)t gives v² = u² + 2as.

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Summary

• The five SUVAT equations apply only when acceleration is constant.
• Always define the positive direction clearly.
• On s-t graphs, gradient = velocity. On v-t graphs, gradient = acceleration and area = displacement.
• For vertical motion, use a = ±g (depending on your sign convention).
• For multi-stage problems, apply SUVAT separately to each stage.