Graphs of motion IB DP Physics Study Notes - 2025 Syllabus
Graphs of motion IB DP Physics Study Notes
Graphs of motion IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
- motion with uniform and non-uniform acceleration
- Motion Graphs
- Motion of a Bouncing Ball
Standard level and higher level: 9 hours
Additional higher level: There is no additional higher level content
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Motion with Uniform and Non-Uniform Acceleration
Uniform Acceleration
- Acceleration \( a \) is constant throughout the motion
- Velocity changes at a constant rate
- Displacement–time graph: curved (parabolic)
- Velocity–time graph: straight line (slope = acceleration)
- Acceleration–time graph: horizontal line
- Equations of motion can be used:
- \( v = u + at \)
- \( s = ut + \frac{1}{2}at^2 \)
- \( v^2 = u^2 + 2as \)
- \( s = \frac{v + u}{2} \cdot t \)
Non-Uniform Acceleration
- Acceleration \( a \) is not constant — it may increase or decrease with time
- Velocity changes at a non-constant rate
- Displacement–time graph: irregular curve
- Velocity–time graph: curved line (slope is not constant)
- Acceleration–time graph: varying (not flat)
- Equations of motion for constant acceleration do not apply
- Need to use calculus:
- Velocity from displacement: \( v = \frac{ds}{dt} \)
- Acceleration from velocity: \( a = \frac{dv}{dt} \)
- Displacement from velocity: \( s = \int v \, dt \)
Example:
A car initially at rest accelerates uniformly at \( 3 \, \text{m/s}^2 \) for \( 8 \, \text{s} \).
Find:
- (a) The final velocity
- (b) The total displacement during this time
▶️ Answer/Explanation
Given:
Initial velocity \( u = 0 \, \text{m/s} \)
Acceleration \( a = 3 \, \text{m/s}^2 \)
Time \( t = 8 \, \text{s} \)
(a) Final velocity
Use: \( v = u + at \)
\( v = 0 + 3 \cdot 8 = 24 \, \text{m/s} \)
(b) Displacement
Use: \( s = ut + \frac{1}{2}at^2 \)
\( s = 0 + \frac{1}{2} \cdot 3 \cdot 8^2 = \frac{3}{2} \cdot 64 = 96 \, \text{m} \)
Final velocity \( = \boxed{24 \, \text{m/s}} \)
Displacement \( = \boxed{96 \, \text{m}} \)
Example:
A car moves along a straight road in two phases:
- Phase 1: It starts from rest and accelerates uniformly at \( 2 \, \text{m/s}^2 \) for \( 10 \, \text{s} \)
- Phase 2: After that, it experiences a velocity-dependent deceleration given by \( a = -0.1v \), where \( v \) is in m/s
Calculate:
- (a) The velocity at the end of Phase 1
- (b) The total distance travelled until the car comes to rest
▶️ Answer/Explanation
(a) Velocity at end of Phase 1
Use \( v = u + at \)
\( v = 0 + 2 \cdot 10 = 20 \, \text{m/s} \)
(b) Distance in Phase 1
Use \( s = ut + \frac{1}{2}at^2 \)
\( s_1 = 0 + \frac{1}{2} \cdot 2 \cdot 10^2 = 100 \, \text{m} \)
Phase 2: Non-uniform acceleration
Given \( a = \frac{dv}{dt} = -0.1v \), this is a separable differential equation.
Step 1: Separate and integrate
\( \frac{dv}{v} = -0.1 \, dt \)
Integrate both sides: \( \int \frac{1}{v} \, dv = \int -0.1 \, dt \)
\( \ln v = -0.1t + C \)
Step 2: Solve for \( v \)
Exponentiate: \( v = Ce^{-0.1t} \)
At \( t = 0 \), \( v = 20 \Rightarrow C = 20 \)
So \( v = 20e^{-0.1t} \)
Step 3: Find distance in Phase 2
Since \( v = \frac{ds}{dt} = 20e^{-0.1t} \)
Integrate to find displacement: \( s_2 = \int 20e^{-0.1t} \, dt \)
\( s_2 = \frac{20}{-0.1}e^{-0.1t} + K = -200e^{-0.1t} + K \)
As \( t \to \infty \), \( v \to 0 \), so motion stops, and \( e^{-0.1t} \to 0 \)
Hence, final value of \( s_2 = -200(0) + K – (-200(1)) = 200 \, \text{m} \)
Total Distance Travelled:
\( s = s_1 + s_2 = 100 + 200 = \boxed{300 \, \text{m}} \)
Motion Graphs
1. Displacement–Time Graphs
- Gradient (slope) represents velocity: \( v = \frac{\Delta s}{\Delta t} \)
- A horizontal line → object is stationary
- A straight line with constant positive slope → constant velocity
- A curved line → changing velocity (i.e., acceleration)
Example:
A cyclist rides in a straight line for 20 minutes, waits for 30 minutes, then returns to her starting point in 15 minutes. This is represented by a displacement-time graph.
Calculate:
- (a) The average velocity for each stage of the journey
- (b) The average velocity for the whole journey
- (c) The average speed for the whole journey
▶️ Answer/Explanation
(a) Average velocity for each stage
- O to A: Time = 20 min, Displacement = 5 km
\( v = \frac{5}{20} = 0.25 \, \text{km/min} = 15 \, \text{km/h} \) - A to B: No displacement change, so \( v = 0 \, \text{km/h} \)
- B to C: Time = 15 min, Displacement = -5 km
\( v = \frac{-5}{15} = -\frac{1}{3} \, \text{km/min} = -20 \, \text{km/h} \)
(b) Average velocity for the whole journey
Total displacement = 0
Average velocity \( = \frac{0}{65} = \boxed{0 \, \text{km/h}} \)
(c) Average speed for the whole journey
Total time = 20 + 30 + 15 = 65 min
Total distance = \( 5 + 5 = 10 \, \text{km} \)
Average speed \( = \frac{10}{65} = \frac{2}{13} \, \text{km/min} \)
Convert to km/h: \( \frac{2}{13} \times 60 = \boxed{9.2 \, \text{km/h}} \) (2 s.f.)
2. Velocity–Time Graphs
- Gradient represents acceleration: \( a = \frac{\Delta v}{\Delta t} \)
- A straight horizontal line → constant velocity
- A sloped line → constant acceleration or deceleration
- Area under the graph gives displacement
Important Cases:
- Positive slope → acceleration
- Negative slope → deceleration
- Line crossing time-axis → change in direction
Example:
A 50-second speed-time graph for a car is shown with 4 labelled sections A, B, C, and D.
Use the graph to calculate:
- (a) The acceleration in each section
- (b) The total distance travelled during the 50 seconds
▶️ Answer/Explanation
(a) Calculating Acceleration (Gradient of Speed-Time Graph)
- Section A:
\( a = \frac{15 – 0}{10 – 0} = \boxed{1.5 \, \text{m/s}^2} \) - Section B:
Flat section → \( a = 0 \, \text{m/s}^2 \) - Section C:
\( a = \frac{25 – 15}{30 – 20} = \boxed{1 \, \text{m/s}^2} \) - Section D:
\( a = \frac{0 – 25}{50 – 30} = \frac{-25}{20} = \boxed{-1.25 \, \text{m/s}^2} \)
So the greatest acceleration is in Section A: \( \boxed{1.5 \, \text{m/s}^2} \)
(b) Calculating Total Distance (Area under Speed-Time Graph)
- Area A (triangle):
\( \frac{1}{2} \cdot 10 \cdot 15 = \boxed{75 \, \text{m}} \) - Area B (rectangle):
\( 10 \cdot 15 = \boxed{150 \, \text{m}} \) - Area C (trapezium):
\( \frac{1}{2} \cdot (15 + 25) \cdot 10 = \boxed{200 \, \text{m}} \) - Area D (triangle):
\( \frac{1}{2} \cdot 20 \cdot 25 = \boxed{250 \, \text{m}} \)
Total Distance:
\( 75 + 150 + 200 + 250 = \boxed{675 \, \text{m}} \)
3. Acceleration–Time Graphs
- Area under the graph gives change in velocity: \( \Delta v = \int a \, dt \)
- Horizontal line above time-axis → constant positive acceleration
- Line at zero → constant velocity (zero acceleration)
- Negative value (line below time-axis) → negative acceleration (deceleration)
Example :
From the acceleration vs time graph given, determine the velocity at \( t = 6 \, \text{s} \), if \( v(0) = 0 \).
▶️ Answer/Explanation
Step 1: Use definition of acceleration
\( a = \frac{dv}{dt} \Rightarrow dv = a \, dt \)
Step 2: Integrate both sides
\( \int dv = \int a \, dt = \int 1.5 \, dt \)
\( v(t) = 1.5t + c \)
Step 3: Apply initial condition
Given \( v(0) = 0 \), so \( c = 0 \)
Therefore, \( v(t) = 1.5t \)
Step 4: Find velocity at \( t = 6 \)
\( v(6) = 1.5 \times 6 = \boxed{9 \, \text{m/s}} \)
Example :
From the acceleration vs time graph given, find the initial velocity of a body if its final velocity is \( 55 \, \text{m/s} \).
▶️ Answer/Explanation
Step 1: Use area under the graph to find change in velocity
Graph has a triangle and a rectangle:
Area \( = \frac{1}{2} \cdot 8 \cdot 6 + 2 \cdot 8 = 24 + 16 = 42 \, \text{m/s} \)
Step 2: Use definition of change in velocity
\( \Delta v = v_f – v_i \Rightarrow 42 = 55 – v_i \)
\( v_i = 55 – 42 = \boxed{13 \, \text{m/s}} \)
Motion of a Bouncing Ball
1. Description of Motion
- The ball is initially dropped from a height with zero initial velocity \( (u = 0) \)
- It accelerates downward due to gravity at \( g = 9.8 \, \text{m/s}^2 \)
- Upon hitting the ground, the velocity becomes zero momentarily, and then the ball rebounds upward
- The rebound height is less than the original height due to energy losses (mostly heat and sound)
- The process repeats with diminishing maximum height on each bounce
2. Graphical Representation
- Displacement–Time graph: shows decreasing peak height with each bounce, resembling a series of upward-opening parabolas
- Velocity–Time graph: consists of straight line segments with alternating positive and negative slopes, with sudden reversals in direction after impact
- Acceleration–Time graph: shows constant acceleration \( -g \) between bounces, and sharp spikes during the very short contact time with the ground
3. Energy Considerations
- On the way down: gravitational potential energy converts to kinetic energy
- On the way up: kinetic energy converts back to potential energy
- Not all energy is recovered due to non-conservative forces (friction, deformation, sound)
- The ball eventually comes to rest due to continual energy loss
4. Realistic Assumptions
- Air resistance is usually neglected for ideal analysis
- Contact time with the ground is short but not zero
- Coefficient of restitution \( e \) describes how “bouncy” the collision is (with \( 0 < e < 1 \))
Example
A ball is dropped from rest and takes \( 0.9 \, \text{s} \) to hit the ground. It rebounds and takes \( 0.6 \, \text{s} \) to reach the maximum height. Find:
- (a) The height from which it was dropped
- (b) The height it reaches after bouncing
▶️ Answer/Explanation
Given:
Downward motion: \( u = 0 \), \( t = 0.9 \, \text{s} \), \( g = 9.8 \, \text{m/s}^2 \)
Rebound time to max height: \( t = 0.6 \, \text{s} \)
(a) Use: \( s = \frac{1}{2}gt^2 \)
\( s = \frac{1}{2} \cdot 9.8 \cdot (0.9)^2 = 4.0 \, \text{m} \)
(b) Use: \( s = \frac{1}{2}gt^2 \) again for rebound
\( s = \frac{1}{2} \cdot 9.8 \cdot (0.6)^2 = 1.764 \, \text{m} \)
Final Answers:
(a) Drop height \( = \boxed{4.0 \, \text{m}} \)
(b) Rebound height \( = \boxed{1.76 \, \text{m}} \)