Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Kinematics
Velocity-time graphs and non-uniform acceleration
- 3 things to know
- Component 1
Overview
Recall questions on the gradient as acceleration and the area as displacement, the shape for uniform acceleration, areas below the axis, finding an instantaneous acceleration from a tangent, and estimating an area under a curve.
What you need to know
Answer each one before you reveal it.
Show by units that the area under a velocity-time graph must be a displacement.
What shape is a velocity-time graph for a body with uniform acceleration?
Questions from this subtopic
Each one with the model answer underneath.
- 1
What physical quantity is given by the gradient of a velocity-time graph?
Model answer
The gradient of a velocity-time graph gives the acceleration of the body.
- 2
What physical quantity is given by the area between a velocity-time graph and the time axis?
Model answer
That area gives the displacement of the body.
- 3
Show by units that the area under a velocity-time graph must be a displacement.
Model answer
The area has units m s⁻¹ × s = m. Those are the units of displacement.
- 4
What shape is a velocity-time graph for a body with uniform acceleration?
Model answer
The graph is a straight line of constant gradient.
- 5
How is an area below the time axis on a velocity-time graph interpreted?
Model answer
An area below the axis represents displacement in the negative direction and is subtracted from the area above.
- 6
A velocity-time graph rises linearly from 4.0 m s⁻¹ at t = 0 to 20.0 m s⁻¹ at t = 8.0 s. Calculate the acceleration and the distance travelled.
Model answer
The acceleration is (20.0 − 4.0)/8.0 = 2.0 m s⁻² and the area is the trapezium ½(4.0 + 20.0) × 8.0 = 96 m. Substituting back, s = ut + ½at² = 4.0 × 8.0 + ½ × 2.0 × 64 = 32 + 64 = 96 m.
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