Uniformly accelerated motion: the equations of kinematics IB DP Physics Study Notes - 2025 Syllabus
Uniformly accelerated motion: the equations of kinematics IB DP Physics Study Notes
Uniformly accelerated motion: the equations of kinematics 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
- the equations of motion for solving problems with uniformly accelerated motion
- $v = u + at$
- $s = ut + \frac{1}{2}at^2$
- $v^2 = u^2 + 2as$
- $s = \frac{u+v}{2}t$
- Determining the Acceleration of Free-Fall Experimentally
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
Equations of Motion
Displacement (\( s \))
The change in position of an object in a particular direction.
Vector quantity, SI unit: meter (m).
Velocity (\( v \))
The rate of change of displacement with time.
Initial velocity: \( u \), Final velocity: \( v \)
\( v = \frac{\text{displacement}}{\text{time}} \)
SI unit: m/s
Acceleration (\( a \))
The rate of change of velocity with time.
\( a = \frac{v – u}{t} \) SI unit: m/s²
Time (\( t \))
Duration of motion, measured in seconds (s).
Assumptions for Equations of Motion
- Motion is in a straight line (1D).
- Acceleration is constant.
- Variables: \( u, v, a, t, s \)
Equation 1: \( v = u + at \)
Derivation:
From the definition of acceleration:
\( a = \frac{v – u}{t} \)
Multiplying both sides by \( t \): \( at = v – u \)
Rearranged: \( \boxed{v = u + at} \)
Equation 2: \( s = ut + \frac{1}{2}at^2 \)
Derivation:
Displacement is the area under the velocity-time graph.
Average velocity: \( v_{\text{avg}} = \frac{u + v}{2} \) \( s = v_{\text{avg}} \cdot t = \frac{u + v}{2} \cdot t \)
Substitute \( v = u + at \): \( s = \frac{u + (u + at)}{2} \cdot t = \frac{2u + at}{2} \cdot t \)
\( \boxed{s = ut + \frac{1}{2}at^2} \)
Equation 3: \( v^2 = u^2 + 2as \)
Derivation:
From Equation 1: \( t = \frac{v – u}{a} \)
Substitute into Equation 2: \( s = u \cdot \frac{v – u}{a} + \frac{1}{2}a \cdot \left(\frac{v – u}{a}\right)^2 \)
\( s = \frac{u(v – u)}{a} + \frac{(v – u)^2}{2a} \)
\( s = \frac{v^2 – u^2}{2a} \Rightarrow \boxed{v^2 = u^2 + 2as} \)
Derivation of \( s = \frac{v + u}{2} \cdot t \)
In uniformly accelerated motion, acceleration is constant, so the velocity-time graph is a straight line.
Average velocity:
\( v_{\text{avg}} = \frac{u + v}{2} \)
Displacement is given by:
\( s = \text{average velocity} \times \text{time} \) \( s = \frac{u + v}{2} \cdot t \)
This formula calculates the displacement \( s \) when the initial velocity \( u \), final velocity \( v \), and time \( t \) are known, under constant acceleration.
\( \boxed{s = \frac{v + u}{2} \cdot t} \)
Summary
Example:
A ball is thrown vertically upwards with an initial velocity of \( u = 20 \, \text{m/s} \). Assume acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \).
Find the maximum height reached by the ball.
▶️ Answer/Explanation
Use the equation \( v^2 = u^2 + 2as \)
At maximum height, final velocity \( v = 0 \)
Acceleration \( a = -g = -9.8 \, \text{m/s}^2 \)
Initial velocity \( u = 20 \, \text{m/s} \)
Substitute values:
\( 0 = (20)^2 + 2(-9.8)h \)
\( 0 = 400 – 19.6h \)
\( h = \frac{400}{19.6} \approx 20.41 \, \text{m} \)
Example:
A car accelerates uniformly from rest to a velocity of \( 30 \, \text{m/s} \) in \( 10 \, \text{s} \), then continues at this velocity for another \( 5 \, \text{s} \), and finally decelerates uniformly to rest in \( 8 \, \text{s} \).
Find the total distance travelled by the car.
▶️ Answer/Explanation
: Acceleration Phase (0 to 10 s)
Given: \( u = 0 \), \( v = 30 \, \text{m/s} \), \( t = 10 \, \text{s} \)
Use: \( s = \frac{u + v}{2} \cdot t \)
\( s_1 = \frac{0 + 30}{2} \cdot 10 = 15 \cdot 10 = 150 \, \text{m} \)
Constant Velocity Phase (10 to 15 s)
Speed = \( 30 \, \text{m/s} \), Time = \( 5 \, \text{s} \)
\( s_2 = 30 \cdot 5 = 150 \, \text{m} \)
Deceleration Phase (15 to 23 s)
Given: \( u = 30 \, \text{m/s} \), \( v = 0 \), \( t = 8 \, \text{s} \)
Use: \( s = \frac{u + v}{2} \cdot t \)
\( s_3 = \frac{30 + 0}{2} \cdot 8 = 15 \cdot 8 = 120 \, \text{m} \)
Total Distance:
\( s = s_1 + s_2 + s_3 = 150 + 150 + 120 = 420 \, \text{m} \)
Determining the Acceleration of Free-Fall Experimentally
Concept:
The acceleration due to gravity, \( g \), is the rate at which objects fall towards Earth under the influence of gravity alone. We can measure it using simple equipment like a falling object, an electronic timer, and a trapdoor mechanism.
Common Method (Trapdoor and Electromagnet):
- A small metal ball is held by an electromagnet above a trapdoor.
- When the electromagnet is turned off, the timer starts, and the ball begins to fall.
- When the ball hits the trapdoor, the timer stops.
Assumption:
Air resistance is negligible, and the motion is uniformly accelerated under gravity.
Equation Used:
Using the equation of motion: \( s = ut + \frac{1}{2}gt^2 \)
Since initial velocity \( u = 0 \), it simplifies to: \( s = \frac{1}{2}gt^2 \)
Rearranged to calculate \( g \): \( g = \frac{2s}{t^2} \)
Plotting Graph (Advanced Method):
- Measure time \( t \) for various drop heights \( s \)
- Plot \( s \) on the y-axis and \( t^2 \) on the x-axis
- Slope of the graph gives \( \frac{g}{2} \), so multiply by 2 to get \( g \)
Precautions:
- Ensure the ball falls vertically without bouncing
- Repeat readings and take an average to reduce random errors
- Use a vacuum if possible to eliminate air resistance
Example:
A student drops a metal ball from a height of \( 1.20 \, \text{m} \). The time measured using an electronic timer is \( 0.495 \, \text{s} \).
Calculate the experimental value of \( g \) from this data.
▶️ Answer/Explanation
Use the equation \( s = \frac{1}{2}gt^2 \)
Rearranged: \( g = \frac{2s}{t^2} \)
Substitute: \( s = 1.20 \, \text{m} \), \( t = 0.495 \, \text{s} \)
\( g = \frac{2 \times 1.20}{(0.495)^2} \approx \frac{2.40}{0.2450} \approx 9.80 \, \text{m/s}^2 \)