Using Differentiation for Kinematics (Edexcel IGCSE Maths B): Revision Note

Exam code: 4MB1

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on

Kinematics

What does kinematics mean?

  • Kinematics is the study of the motion of an object

    • It is a branch of Physics

  • Objects are called particles

    • They are modelled as single moving points

  • Over time, the particles move and

    • can be at different distances from a fixed origin (displacement)

    • can move with different speeds in different directions (velocity)

    • can speed up or slow down (acceleration)

What is displacement?

  • The displacement of an object is how far away it is from a fixed origin

    • It can be positive (in front of the origin)

    • or negative (behind the origin)

  • Do not confuse displacement with distance

    • Distance is always positive!

    • Displacement can have a ± sign

  • Displacement is given the letter s in kinematics

    • Do not confuse this letter for speed!

    • Displacement is measured in metres

Diagrams showing displacement examples: one particle 4 metres right of point 0, another 5 metres left of point 0. Text explains displacement in each case.

What is a displacement function?

  • The displacement of an object, s metres, can be written as a function of time, t seconds

    • s=f(t)

  • Substitute a value of time in to find its displacement at that time

    • For example, s=2tt2+1

      • Initially, t=0 gives s=2×002+1=1 (1 metre in front of the origin)

      • After 3 seconds, t=3 gives s=2×332+1=2 (2 metres behind the origin)

What is the velocity and how do I find it?

  • Velocity is the speed and direction of an object

    • It is positive if moving forwards

    • It is negative if moving backwards

    • Do not confuse velocity and speed

      • Speed is always positive!

  • To find the velocity of an object, v metres per second, differentiate its displacement function

    • v=dsdt

  • For example, if s=t34t2+2t then v=3t28t+2 (by differentiation)

    • The initial velocity (t=0) is v=3×028×0+2=2 ms-1

    • The velocity after 1 second (t=1) is v=3×128×1+2=3 ms-1

      • Its speed is 3 ms-1

  • If a velocity is zero at any point in time, it is said to be at instantaneous rest

    • It is stationary (not moving) at that instant in time

      • but not stationary all the time

    • To find the times at which the particle is at rest, set v=0 and solve to find t

What is the acceleration and how do I find it?

  • Acceleration is rate at which the velocity changes

    • It is positive if speeding up (when moving forwards)

    • It is negative if slowing down (when moving forwards)

      • A negative acceleration is also called a deceleration

      • The magnitude of acceleration is always positive

  • To find the acceleration of an object, a metres per second per second, differentiate its velocity function

    • a=dvdt

  • For example, if v=3t28t+2 then a=6t8

    • You can substitute times in to find accelerations

  • It the acceleration is always zero then the particle moves at a constant speed

How do I find the acceleration from the displacement?

  • You differentiate displacement to get velocity, then differentiate velocity to get acceleration

    • So you differentiate displacement twice to get acceleration

Diagram showing the relationship between displacement (s), velocity (v), and acceleration (a), with differentiation denoted as ds/dt for velocity and dv/dt for acceleration.

Examiner Tips and Tricks

  • Harder exam questions may jump back and forth between displacement, velocity and acceleration

    • so make sure you use the labels s=..., v=... and a=... to make your working clear

Worked Example

A particle moves along a straight line.

The displacement of the particle from a fixed point, O, on the line at time t seconds is s metres, where

s=t36t23

(a) Find the initial distance of the particle from O.

Answer:

Initial means t=0
Substitute t=0 into s to find the initial displacement

s=036×023=3

Distance is always positive, so convert -3 into 3

The particle is initially at a distance of 3 metres from O

(b) Find an expression for the velocity, v ms-1, at time t seconds.

Answer:

To find the velocity, differentiate the displacement

v=dsdt=3t212t

This is an expression for the velocity in terms of time, t

v=3t212t ms-1

(c) Find how long, after t=0, it takes for the particle to come to rest.

Answer:

The particle is at rest when v=0
Set v=0 and solve to find t

0=3t212t0=t24t0=t(t4)

t=0 or t=4

After t=0 the next point of rest is t=4

After t=0, it takes 4 seconds for the particle to come to rest

(d) Find the time at which the particle is decelerating at 3 ms-2.

Answer:

A deceleration of 3 means an acceleration of -3
Differentiate the velocity function to find acceleration

a=dvdt=6t12

Set a=3 and solve for t

6t12=36t=9t=1.5

The particle is decelerating at 3 ms-2 at 1.5 seconds

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Mark Curtis

Author: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.