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CIE A-Level Maths Study Notes

3.2.3 Calculus in Kinematics

Calculus, with its two main operations - differentiation and integration - plays an essential role in the study of kinematics, particularly in understanding motion along a straight line. This section delves into how differentiation and integration can be applied to solve problems related to displacement, velocity, and acceleration.

Application of Differentiation in Kinematics

  • Differentiation helps calculate how fast things change. In kinematics, it's used for finding velocity and acceleration from how position changes over time.
  • Velocity from Displacement:
    • Velocity is how fast position changes.
    • It's calculated as the rate of change of displacement s(t)s(t) over time tt: v(t)=dsdtv(t) = \frac{ds}{dt} .
  • Acceleration from Velocity:
    • Acceleration is how fast velocity changes.
    • It's the rate of change of velocity over time, or the second rate of change of displacement: a(t)=dvdt=d2sdt2a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}.
  • Integration in Kinematics:
    • Integration is the reverse of differentiation. It's used to find displacement from velocity.
    • Displacement ss is the accumulated total of velocity v(t)v(t) over time: s(t)=v(t)dts(t) = \int v(t) \, dt.
relationship between acceleration , velocity and displacement

Image courtesy of webtemplate

Examples

Problem 1: Calculating Displacement from Velocity

Question: Find the displacement of a particle moving in a straight line with velocity v(t)=3t22tv(t) = 3t^2 - 2t from t=1t = 1 to t=4t = 4 seconds.

Solution:

  • Understand: Displacement is the integral of velocity over time.
  • Set Up Integral: s(t)=(3t22t)dts(t) = \int (3t^2 - 2t) \, dt.
  • Integration Limits: From t=1t = 1 to t=4t = 4.
  • Integrate: (3t22t)dt=t3t2\int (3t^2 - 2t) \, dt = t^3 - t^2 .
  • Evaluate: s=[(4342)(1312)]s = [(4^3 - 4^2) - (1^3 - 1^2)] .
  • Result: s=48s = 48 units.
Velocity and Displacement

Problem 2: Finding Velocity from Acceleration

Question: Find the velocity after 5 seconds for a particle starting from rest with acceleration a(t)=6ta(t) = 6t .

Solution:

  • Understand: Velocity is the integral of acceleration over time.
  • Initial Condition: Particle starts from rest, so initial velocity v(0)=0v(0) = 0 .
  • Set Up Integral: v(t)=6tdt+v(0)v(t) = \int 6t \, dt + v(0) .
  • Integrate: 6tdt=3t2\int 6t \, dt = 3t^2 .
  • Velocity Function: v(t)=3t2v(t) = 3t^2 .
  • Find Velocity at t=5t=5: v(5)=75v(5) = 75 units.
Velocity of Particle Overtime
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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