IB Mathematics AA SL Kinematic problems involving displacement, velocity and acceleration Study Notes - New Syllabus
IB Mathematics AA SL Kinematic problems involving displacement, velocity and acceleration Study Notes
LEARNING OBJECTIVE
- Kinematic problems
Key Concepts:
- Kinematic involving calculus
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
KINEMATICS IN CALCULUS (MOTION IN ONE DIMENSION)
Fundamental Concepts in Kinematics
Displacement \( s(t) \): Represents an object’s position along a line at any time \( t \).
Velocity \( v(t) \): The rate at which displacement changes over time.
Acceleration \( a(t) \): Describes how velocity itself changes with respect to time.
Key Calculus Formulas
- \(\boxed{ v(t) = \frac{ds}{dt} }\) — Velocity as the first derivative of position.
- \( \boxed{ a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} }\) — Acceleration as the derivative of velocity or second derivative of position.
- \(\boxed{ a = v \cdot \frac{dv}{ds} }\) — Chain rule form relating acceleration, velocity, and displacement.
| Displacement | Velocity | Acceleration |
|---|---|---|
| The change in position of an object in a specific direction. | The rate of change of displacement with respect to time. | The rate of change of velocity with respect to time. |
| Vector | Vector | Vector |
| Has both magnitude and direction. | Can be positive, negative, or zero depending on direction. | Can be positive (speeding up), negative (slowing down), or zero (constant velocity). |
| \( \Delta x = x_{\text{final}} – x_{\text{initial}} \) | \( v = \dfrac{\text{displacement}}{\text{time}} \) | \( a = \dfrac{\text{change in velocity}}{\text{time}} \) |
| Can be zero even if there is movement (e.g., returning to starting point). | Zero when displacement is constant. | Zero when velocity is constant (uniform motion). |
| Meters (m) | Meters per second (m/s) | Meters per second squared (m/s²) |
Example:
An object moves along a line such that its velocity is described by \( v(t) = 4t^2 – 16t + 12 \). Determine the following:
- The displacement of the object from \( t = 1 \) to \( t = 4 \)
- The total distance travelled over the same interval
- The value of acceleration at \( t = 3 \)
▶️ Answer/Explanation
Displacement
Integrate the velocity function over [1, 4]:
$ s = \int_1^4 (4t^2 – 16t + 12) \, dt = \left[ \frac{4}{3}t^3 – 8t^2 + 12t \right]_1^4 $
$ = \left(\frac{256}{3} – 128 + 48\right) – \left(\frac{4}{3} – 8 + 12\right) = \left(\frac{256 – 384 + 144}{3}\right) – \left(\frac{4 + 12 – 24}{3}\right) = \frac{16}{3} $
Displacement $= \rm{\frac{16}{3} \text{ units}}$
Total Distance Travelled
Find when \( v(t) = 0 \):
$ 4t^2 – 16t + 12 = 0 \Rightarrow t = 1, 3 $
Break into intervals: [1, 3] and [3, 4]
$ \text{Distance on } [1,3] = \int_1^3 (4t^2 – 16t + 12) dt = 0 \quad \text{(since it returns to starting point)} $
$ \text{Distance on } [3,4] = \int_3^4 (4t^2 – 16t + 12) dt = \left[\frac{4}{3}t^3 – 8t^2 + 12t\right]_3^4 $
$ = \left(\frac{256}{3} – 128 + 48\right) – \left(\frac{108}{3} – 72 + 36\right) = \frac{16}{3} $
Total distance $= \rm{\frac{16}{3} \text{ units}}$
Acceleration at \( t = 3 \)
Differentiate velocity:
$ a(t) = \frac{dv}{dt} = 8t – 16 \Rightarrow a(3) = 8(3) – 16 = \rm{8} $
DISPLACEMENT AND DISTANCE
Displacement from Velocity
If the velocity function \( v(t) \) is known, the displacement of a moving object over a time interval \([t_1, t_2]\) can be computed as:
\(\boxed{ \text{Displacement} = \int_{t_1}^{t_2} v(t)\, dt }\)
This integral gives the net change in position, which includes direction – so the result may be negative if the particle moves backward.
Example A particle moves with velocity \( v(t) = 2t^2 – 8 \) for \( 1 \leq t \leq 5 \). Find the displacement over this interval. ▶️Answer/ExplanationSolution: $ \int_1^5 (2t^2 – 8)\, dt = \left[\frac{2}{3}t^3 – 8t\right]_1^5 = \left(\frac{250}{3} – 40\right) – \left(\frac{2}{3} – 8\right) $ $ = \left(\frac{250 – 120}{3}\right) – \left(\frac{2 + 24}{3}\right) = \frac{130}{3} – \frac{26}{3} = \frac{104}{3} $ Displacement = \(\rm{\frac{104}{3}}\) units |
Total Distance Travelled
The total distance travelled is the sum of all movements, regardless of direction. To compute this, we take the integral of the absolute value of velocity:
\(\boxed{ \text{Total distance} = \int_{t_1}^{t_2} |v(t)|\, dt} \)
Unlike displacement, this value is always non-negative and reflects the total path length covered by the object.
Example A particle’s velocity is given by \( v(t) = t^2 – 6t + 5 \) on the interval \( 0 \leq t \leq 5 \). Find the total distance travelled. ▶️Answer/ExplanationFactor: \( v(t) = (t – 1)(t – 5) \) Velocity changes sign at \( t = 1 \) and \( t = 5 \). Split interval: $ \int_0^1 v(t)\, dt + \int_1^5 |v(t)|\, dt = \int_0^1 (t^2 – 6t + 5) dt + \int_1^5 -(t^2 – 6t + 5) dt $ $ = \left[\frac{1}{3}t^3 – 3t^2 + 5t\right]_0^1 + \left[-\left(\frac{1}{3}t^3 – 3t^2 + 5t\right)\right]_1^5 $ First part: \( \frac{1}{3} – 3 + 5 = \frac{7}{3} \) \( f(5) = \frac{125}{3} – 75 + 25 = \frac{125 – 225 + 75}{3} = -\frac{25}{3} \), Total distance = \( \frac{7}{3} + \left|\left(-\frac{25}{3} – \frac{7}{3}\right)\right| = \frac{7}{3} + \frac{32}{3} = \rm{\frac{39}{3} = 13} \) units |
SPEED VS VELOCITY
Velocity is a vector quantity – it has both magnitude and direction.
Speed is the magnitude of velocity:
$ \boxed{\text{Speed} = |v(t)|} $
Speed is always non-negative, whereas velocity can be negative depending on direction.
| Speed | Velocity |
|---|---|
| The rate at which an object covers distance, regardless of direction. | The rate at which an object changes its position in a specific direction. |
| Scalar | Vector |
| \( v = \dfrac{\text{distance}}{\text{time}} \) | \( v = \dfrac{\text{displacement}}{\text{time}} \) |
| Direction does not affect speed. | Changes in direction result in changes in speed. |
| The speed is zero only when the object is stationary. | It can be zero even when there is motion when the net displacement is zero. |
| Meters per second (m/s), kilometers per hour (km/h), miles per hour (mph), etc. | Same as speed, but with direction: 60 km/h north. |
Example A cyclist travels with a velocity of \( v(t) = -12 \) km/h for \( 0 \leq t \leq 2 \). What is the cyclist’s speed? ▶️Answer/ExplanationSpeed is the magnitude of velocity: $|v(t)| = |-12| = 12$ km/h |
