CIE AS/A Level Maths-4.2 Kinematics of motion in a straight line - Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-4.2 Kinematics of motion in a straight line- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-4.2 Kinematics of motion in a straight line- Study Notes
Key Concepts:
- Scalar and Vector Quantities in Kinematics (1D Motion)
- Graphical Representation of Motion (1D)
- Calculus in 1D Kinematics
- Equations of Motion (SUVAT equations)
Scalar and Vector Quantities in Kinematics (1D Motion)
Scalar and Vector Quantities in Kinematics (1D Motion)
- Scalars: Quantities that have only magnitude (size), no direction.
- Vectors: Quantities that have both magnitude and direction.
Distance and Speed (Scalars)
- Distance: The total length of the path travelled by a particle, irrespective of direction.
Distance \(\geq 0\), and it always increases as motion continues.
- Speed: The rate of change of distance with respect to time.
\(\text{Speed} = \dfrac{\text{Distance travelled}}{\text{Time taken}}\)
Units: m/s (metres per second)
Displacement, Velocity and Acceleration (Vectors)
- Displacement: The straight-line change in position of a particle from its initial position to its final position, including direction.
Displacement may be positive or negative, depending on direction chosen as positive.
- Velocity: The rate of change of displacement with respect to time.
\(\text{Velocity} = \dfrac{\text{Displacement}}{\text{Time}}\)
Velocity can be positive (motion in chosen positive direction) or negative (motion in opposite direction).
- Acceleration: The rate of change of velocity with respect to time.
- Acceleration: The rate of change of velocity with respect to time.
\(\text{Acceleration} = \dfrac{\Delta v}{\Delta t}\)
Acceleration can be positive (speeding up in positive direction) or negative (slowing down or accelerating in negative direction).
Deceleration
- Deceleration is not a separate vector quantity, but refers to a decrease in the magnitude of velocity.
- In 1D motion, deceleration corresponds to acceleration acting in the opposite direction to velocity.
Example:
A car travels in a straight line. It moves 100 m east in 20 s, then 60 m west in 10 s.
▶️ Answer/Explanation
Step 1: Distance travelled
Total distance = \(100 + 60 = 160 \, \text{m}\)
Step 2: Displacement
Take east as positive. Net displacement = \(+100 – 60 = +40 \, \text{m}\) east.
Step 3: Average speed
\(\text{Average speed} = \dfrac{160}{30} = 5.33 \, \text{m/s}\)
Step 4: Average velocity
\(\text{Average velocity} = \dfrac{40}{30} = 1.33 \, \text{m/s east}\)
Final Answer:
\(\boxed{\text{Distance = 160 m, Displacement = 40 m east, Average speed = 5.33 m/s, Average velocity = 1.33 m/s east}}\)
Graphical Representation of Motion (1D)
Graphical Representation of Motion (1D)
Displacement–Time Graphs
- The gradient of a displacement–time graph at any point gives the velocity of the object.
- A straight line with constant gradient → uniform velocity.
- A horizontal line → zero velocity (object at rest).
- A curved line → changing velocity (acceleration or deceleration).
Velocity–Time Graphs
- The gradient of a velocity–time graph at any point gives the acceleration.
- A horizontal line → constant velocity.
- A sloping line → uniform acceleration or deceleration.
- The area under the velocity–time graph between two times gives the displacement during that interval.
Key Interpretations
- Gradient of displacement–time graph = velocity.
- Gradient of velocity–time graph = acceleration.
- Area under velocity–time graph = displacement.
Example:
A particle moves in a straight line. Its velocity–time graph is a straight line increasing from \(0 \, \text{m/s}\) at \(t = 0\) to \(10 \, \text{m/s}\) at \(t = 5 \, \text{s}\).
▶️ Answer/Explanation
Step 1: Identify acceleration
Gradient = \(\dfrac{10 – 0}{5 – 0} = 2 \, \text{m/s}^2\).
Step 2: Find displacement from area
Area under graph = area of triangle = \(\dfrac{1}{2} \times 5 \times 10 = 25 \, \text{m}\).
Step 3: Interpret motion
The object starts from rest, accelerates uniformly at \(2 \, \text{m/s}^2\), and travels a displacement of \(25 \, \text{m}\) in \(5 \, \text{s}\).
Final Answer:
\(\boxed{\text{Acceleration = 2 m/s}^2, \; \text{Displacement = 25 m}}\)
Calculus in 1D Kinematics
Calculus in 1D Kinematics
Displacement \(s(t)\), velocity \(v(t)\), acceleration \(a(t)\) are functions of time \(t\).
- Definitions (differentiation with respect to time):
\(v(t) = \dfrac{ds}{dt}, \qquad a(t) = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}\)
- Inverse process (integration with respect to time):
\(s(t) = \int v(t)\,dt + C_1, \qquad v(t) = \int a(t)\,dt + C_2\)
- Use initial/ boundary conditions (e.g. \(s(0)\), \(v(0)\), values at given \(t\)) to determine constants.
- Areas under graphs: area under a \(v\)-\(t\) graph gives displacement; area under an \(a\)-\(t\) graph gives change in velocity.
Differentiation tools you’ll use (Paper 1)
- Power rule: \(\dfrac{d}{dt}(t^n)=nt^{n-1}\) for integer \(n\).
- Exponentials/trig (when used): \(\dfrac{d}{dt}(e^{kt})=ke^{kt}\), \(\dfrac{d}{dt}(\sin kt)=k\cos kt\), \(\dfrac{d}{dt}(\cos kt)=-k\sin kt\).
- Basic product/chain rules as needed for simple forms.
Integration tools you’ll use (Paper 1)
- Power rule: \(\int t^n\,dt=\dfrac{t^{n+1}}{n+1}+C\) for \(n\neq -1\).
- \(\int e^{kt}\,dt=\dfrac{1}{k}e^{kt}+C\); \(\int \sin kt\,dt=-\dfrac{1}{k}\cos kt + C\); \(\int \cos kt\,dt=\dfrac{1}{k}\sin kt + C\).
Workflow checklist
- If \(s(t)\) is given → differentiate once for \(v(t)\), twice for \(a(t)\).
- If \(a(t)\) is given → integrate to get \(v(t)\), then integrate again to get \(s(t)\); use conditions to find constants.
- If \(v(t)\) is given → integrate for \(s(t)\), differentiate for \(a(t)\) if required.
Example
A particle moves on a line with \(s(t)=3t^3-10t^2+4t-7\) (metres). Find \(v(t)\) and \(a(t)\). Find the time when the particle is instantaneously at rest.
▶️ Answer/Explanation
Differentiate:
\(v(t)=\dfrac{ds}{dt}=9t^2-20t+4\)
\(a(t)=\dfrac{dv}{dt}=18t-20\)
Instantaneously at rest \(\Rightarrow v(t)=0\):
\(9t^2-20t+4=0 \Rightarrow t=\dfrac{20\pm\sqrt{400-144}}{18}=\dfrac{20\pm\sqrt{256}}{18}=\dfrac{20\pm16}{18}\)
\(t=\dfrac{36}{18}=2\ \text{s}\) or \(t=\dfrac{4}{18}=\dfrac{2}{9}\ \text{s}\)
\(\boxed{v(t)=9t^2-20t+4,\ a(t)=18t-20,\ \text{rest at } t=\dfrac{2}{9}\text{ s and }2\text{ s}}\)
Example
A particle has acceleration \(a(t)=6t-4\ \text{m s}^{-2}\). Given \(v(0)=3\ \text{m s}^{-1}\) and \(s(0)=2\ \text{m}\), find \(v(t)\) and \(s(t)\).
▶️ Answer/Explanation
Integrate \(a\) to get \(v\):
\(v(t)=\int (6t-4)\,dt=3t^2-4t+C_1\)
Use \(v(0)=3\): \(3=C_1 \Rightarrow v(t)=3t^2-4t+3\).
Integrate \(v\) to get \(s\):
\(s(t)=\int (3t^2-4t+3)\,dt=t^3-2t^2+3t+C_2\)
Use \(s(0)=2\): \(2=C_2 \Rightarrow s(t)=t^3-2t^2+3t+2\).
\(\boxed{v(t)=3t^2-4t+3,\quad s(t)=t^3-2t^2+3t+2}\)
Example
A particle has velocity \(v(t)=5+2t\ \text{m s}^{-1}\) for \(0\le t\le 6\). Find
(i) displacement in the first 6 s,
(ii) the acceleration, and
(iii) the time when displacement from \(t=0\) first reaches \(60\ \text{m}\).
▶️ Answer/Explanation
(i) Displacement is the integral of \(v\):
\(s(6)-s(0)=\int_0^{6}(5+2t)\,dt=\left[5t+t^2\right]_0^{6}=30+36=66\ \text{m}\)
(ii) Acceleration is \(a=\dfrac{dv}{dt}=2\ \text{m s}^{-2}\) (constant).
(iii) Find \(s(t)-s(0)=\int_0^{t}(5+2u)\,du=5t+t^2\). Set \(5t+t^2=60\):
\(t^2+5t-60=0 \Rightarrow t=\dfrac{-5+\sqrt{25+240}}{2}=\dfrac{-5+\sqrt{265}}{2}\approx 5.64\ \text{s}\)
\(\boxed{\text{Displacement }66\ \text{m},\ a=2\ \text{m s}^{-2},\ t\approx 5.64\ \text{s}}\)
Example
A ball is projected vertically upwards with speed \(u=18\ \text{m s}^{-1}\). Take upward as positive and \(g=9.8\ \text{m s}^{-2}\). Find \(v(t)\), \(s(t)\) (with \(s(0)=0\)), the maximum height, and the time of return to \(s=0\).
▶️ Answer/Explanation
Acceleration is constant downward: \(a=-g=-9.8\).
\(v(t)=\int a\,dt=-9.8t+C,\ \ v(0)=18\Rightarrow C=18\Rightarrow v(t)=18-9.8t\)
\(s(t)=\int v\,dt=\int (18-9.8t)\,dt=18t-4.9t^2 + K,\ \ s(0)=0\Rightarrow K=0\)
Maximum height when \(v=0\): \(18-9.8t=0 \Rightarrow t=\dfrac{18}{9.8}\approx 1.837\ \text{s}\).
\(s_{\max}=18(1.837)-4.9(1.837)^2\approx 16.5\ \text{m}\)
Return to ground when \(s=0\) (other than \(t=0\)):
\(18t-4.9t^2=0\Rightarrow t(18-4.9t)=0\Rightarrow t=\dfrac{18}{4.9}\approx 3.673\ \text{s}\)
\(\boxed{v(t)=18-9.8t,\ s(t)=18t-4.9t^2,\ s_{\max}\approx 16.5\ \text{m},\ t_{\text{return}}\approx 3.67\ \text{s}}\)
Equations of Motion (SUVAT equations)
Equations of Motion (SUVAT equations)
When a particle moves in a straight line with constant acceleration, the following equations are used. The variables are
- \(s\): displacement (m)
- \(u\): initial velocity (m/s)
- \(v\): final velocity (m/s)
- \(a\): constant acceleration (m/s\(^2\))
- \(t\): time (s)
1. \(v = u + at\)
2. \(s = ut + \dfrac{1}{2}at^2\)
3. \(s = vt – \dfrac{1}{2}at^2\)
4. \(s = \dfrac{(u+v)}{2}t\)
5. \(v^2 = u^2 + 2as\)
Key Notes:
- These equations are valid only when acceleration \(a\) is constant.
- They apply to motion in a straight line (1D motion).
- They are interrelated, so depending on the given information, different forms are used.
Example:
A car accelerates uniformly from rest with acceleration \(2 \, \text{m/s}^2\) for \(10 \, \text{s}\). Find:
- (a) its final velocity
- (b) the total displacement covered
▶️ Answer/Explanation
Step 1: Identify known values: \(u = 0 \, \text{m/s}, \, a = 2 \, \text{m/s}^2, \, t = 10 \, \text{s}\).
Step 2 (final velocity): Use \(v = u + at\). \(v = 0 + (2)(10) = 20 \, \text{m/s}\).
Step 3 (displacement): Use \(s = ut + \dfrac{1}{2}at^2\). \(s = 0(10) + \dfrac{1}{2}(2)(10^2)\). \(s = 100 \, \text{m}\).
Final Answer:
Final velocity = \(\boxed{20 \, \text{m/s}}\)
Displacement = \(\boxed{100 \, \text{m}}\)
Example:
Two particles \(A\) and \(B\) move along the same straight line in the same direction.
- At \(t=0\), \(A\) is at the origin \(O\) with velocity \(4\ \text{m s}^{-1}\) and constant acceleration \(1.5\ \text{m s}^{-2}\).
- At \(t=0\), \(B\) is \(120\ \text{m}\) ahead of \(A\), moving with velocity \(12\ \text{m s}^{-1}\), and decelerates uniformly at \(0.8\ \text{m s}^{-2}\).
1) Determine if and when \(A\) catches \(B\), and the position where this happens.
2) Find \(B\)’s speed at the instant of being caught and check whether \(B\) had already come to rest before being caught.
3) Find the initial separation that would make \(A\) catch \(B\) exactly when \(B\) comes to rest.
▶️ Answer/Explanation
Step 1: Write equations of motion
For \(A\): \(u_A=4,\ a_A=1.5\)
\(x_A(t) = 4t + 0.75t^2\)
For \(B\): initial \(=120,\ u_B=12,\ a_B=-0.8\)
\(x_B(t) = 120 + 12t – 0.4t^2\)
Step 2: Catching time
Set \(x_A=x_B\):
\(4t+0.75t^2 = 120 + 12t – 0.4t^2\)
\(\Rightarrow 1.15t^2 – 8t – 120 = 0\)
\(t = \dfrac{8+\sqrt{616}}{2(1.15)} \approx 14.26\ \text{s}\)
Position: \(x_A(14.26) \approx 209.7\ \text{m}\)
\(\boxed{t \approx 14.26\ \text{s},\ x \approx 209.7\ \text{m}}\)
Step 3: Speed of \(B\) then
\(v_B(t) = 12 – 0.8t\)
At \(t=14.26\): \(v_B \approx 0.59\ \text{m s}^{-1}\)
Stop time: \(t_{stop} = \dfrac{12}{0.8} = 15\ \text{s}\)
Since \(14.26 < 15\), \(B\) has not stopped yet.
\(\boxed{v_B \approx 0.59\ \text{m s}^{-1}\ \text{(still moving)}}\)
Step 4: Separation for meeting at stop
If they meet at \(t=15\):
\(x_A(15)=228.75\)
\(x_B(15)=D+90\)
\(D+90=228.75 \Rightarrow D=138.75\ \text{m}\)
\(\boxed{138.75\ \text{m}}\)