IBDP Physics SL&HL: A.1 Kinematics Study Notes -First assessment 2025

IB DP PHYSICS 2025 SL&HL – Study notes- All Topics

Topic 1.1 Displacement , Distance , Speed and Velocity

Topic 1 Weightage : XX

Notes for 1.1 Displacement , Distance , Speed and Velocity

  • Determining instantaneous and average values for velocity, speed and acceleration

1.1 Displacement, distance, speed and velocity

Kinematics deals with the concepts that are needed to describe motion.

Dynamics deals with the effect that forces have on motion.

Together, kinematics and dynamics form the branch of physics known as Mechanics.

Displacement :
  1. Displacement may be positive, negative or zero but distance is always positive.
  2. Displacement is not affected by the shift of the coordinate axes.
  3. Displacement of an object is independent of the path followed by the object but distance depends upon path.
  4. Displacement and distance both have same unit as that of length i.e. meter
  5. For a moving body distance always increases with time
  6. For a body undergoing one dimensional motion, in the same direction distance = | displacement |. For all other motion distance > | displacement |.

Speed and Velocity

Average speed is the distance traveled divided by the time required to cover the distance.

\text { Average speed }=\frac{\text { Distance }}{\text { Elapsed time }}

SI units for speed: meters per second \((\mathrm{m} / \mathrm{s})\) 

Example 1 Distance Run by a Jogger

How far does a jogger run in 1.5 hours ( \(5400 \mathrm{~s}\) ) if his average speed is \(2.22 \mathrm{~m} / \mathrm{s}\) ?



& \text { Average speed }=\frac{\text { Distance }}{\text { Elapsed time }} \\
& \text { Distance }=(\text { Average speed })(\text { Elapsed time }) \\
& =(2.22 \mathrm{~m} / \mathrm{s})(5400 \mathrm{~s})=12000 \mathrm{~m}

Average velocity is the displacement divided by the elapsed time.

\text { Average velocity }=\frac{\text { Displacement }}{\text { Elapsed time }} \\
\overrightarrow{\overrightarrow{\mathbf{v}}}=\frac{\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{x}}_o}{t-t_o}=\frac{\Delta \overrightarrow{\mathbf{x}}}{\Delta t}

Example 2 The World’s Fastest Jet-Engine Car

Andy Green in the car ThrustSSC set a world record of \(341.1 \mathrm{~m} / \mathrm{s}\) in 1997. To establish such a record, the driver makes two runs through the course, one in each direction, to nullify wind effects. From the data, determine the average velocity for each run.



\(\begin{gathered}(a): \overline{\overrightarrow{\mathbf{v}}}=\frac{\Delta \overrightarrow{\mathbf{x}}}{\Delta t}=\frac{+1609 \mathrm{~m}}{4.740 \mathrm{~s}}=+339.5 \mathrm{~m} / \mathrm{s} \\(b): \overline{\overrightarrow{\mathbf{v}}}=\frac{\Delta \overrightarrow{\mathbf{x}}}{\Delta t}=\frac{-1609 \mathrm{~m}}{4.695 \mathrm{~s}}=-342.7 \mathrm{~m} / \mathrm{s}\end{gathered}\)

The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.
\overrightarrow{\mathbf{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathbf{x}}}{\Delta t}


The notion of acceleration emerges when a change in velocity is combined with the time during which the change occurs.


\overline{\overrightarrow{\mathbf{a}}}=\frac{\overrightarrow{\mathbf{v}}-\overrightarrow{\mathbf{v}}_o}{t-t_o}=\frac{\Delta \overrightarrow{\mathbf{v}}}{\Delta t}

Example 3 Acceleration and Increasing Velocity

Determine the average acceleration of the plane.


\overrightarrow{\mathbf{v}}_o=0 \mathrm{~m} / \mathrm{s} \quad \overrightarrow{\mathbf{v}}=260 \mathrm{~km} / \mathrm{h} \quad t_o=0 \mathrm{~s} \quad t=29 \mathrm{~s} \\
\overline{\overrightarrow{\mathbf{a}}}=\frac{\overrightarrow{\mathbf{v}}-\overrightarrow{\mathbf{v}}_o}{t-t_o}=\frac{260 \mathrm{~km} / \mathrm{h}-0 \mathrm{~km} / \mathrm{h}}{29 \mathrm{~s}-0 \mathrm{~s}}=+9.0 \frac{\mathrm{km} / \mathrm{h}}{\mathrm{s}}

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