A.1 KinematicsIB Physics SL: Revision notes

Motion is described using position (where a body is relative to a chosen origin), velocity and acceleration. The displacement is the change in position: a vector with both magnitude and direction. The distance travelled is the total length of the path, a scalar.

A runner who completes a full lap of a track has travelled 400 m but has zero displacement. After half a lap of a circular track, the displacement is the diameter, not the distance run.

Velocity is the rate of change of position (displacement ÷ time); speed is its magnitude. Acceleration is the rate of change of velocity.

• Average values use the whole interval: average speed = total distance ÷ total time; average velocity = displacement ÷ total time; average acceleration = change in velocity ÷ time.
• Instantaneous values apply at a single moment. They are found from the gradient of the tangent to a displacement–time curve (for velocity) or a velocity–time curve (for acceleration). From tabulated data, estimate them using a short interval centred on the instant.

The area under a velocity–time graph is the displacement; the area under an acceleration–time graph is the change in velocity.

When the acceleration is uniform (constant), four equations link displacement s, initial velocity u, final velocity v, acceleration a and time t:

• $s = \frac{u + v}{2}t$
• $v = u + at$
• $s = ut + \frac{1}{2}at^2$
• $v^2 = u^2 + 2as$

Choose the equation that contains the three quantities you know and the one you want. Take one direction as positive and give every vector quantity a sign.

With non-uniform acceleration these equations are not valid. Instead, work from graphs or data: find displacement from the area under the v–t graph (e.g. the trapezium method) and acceleration from gradients.

A projectile moves under gravity alone once launched. Resolve the launch velocity u at angle θ into components:

• horizontal: $u_x = u\cos\theta$ — constant, because no horizontal force acts, so $x = u_x t$
• vertical: $u_y = u\sin\theta$ — changes with uniform acceleration g downwards, so use the equations of motion

The two components are independent and are linked only by the time t. At the top of the path the vertical velocity is zero but the horizontal velocity is not. On level ground the path is a symmetrical parabola, the time of flight is $2u_y/g$ and the range is $u_x$ × time of flight.

Fluid resistance (drag) acts opposite to the velocity and increases with speed. For a projectile it:

• reduces the horizontal velocity, so the range decreases
• lowers the maximum height, because drag adds to gravity on the way up
• makes the trajectory asymmetric, with a steeper descent than ascent
• changes the time of flight (usually slightly shorter overall)
• means the acceleration is no longer constant or purely vertical

A body falling through a fluid speeds up until the drag equals its weight. The resultant force is then zero, so it stops accelerating and falls at its terminal speed.