S

t

v

tGradient of ST graph =

Gradient of a VT graph =

Area under a VT graph =

Velocity

Acceleration

Displacement

Variable Acceleration

• Know how displacement, velocity, and acceleration are linked by calculus

• Understand how to use calculus to find equations

• Be able to add something here... It’s alright Kim, no one reads this. Hey, Kim isn’t even your name.

General Motion

• When acceleration is not constant we can use calculus to help use link a, v and s

• We can differentiate when we would have considered gradients

• We can integrate when we would have considered areas.

• To use this method you need an equation as a function of time.

Differentiation

Example:s = 3t5 + 2t3 + t + 5 find v and a when t = 1

1615 24 ttv

tta 1260 3

122 msv

272 msa

tfs tfdt

dsv tf

dt

dva

IntegrationIf a = f(t)

If v = g(t)

Example:a = 2t3 + 6 find the change in velocity between t =0 and t = 2

dttfv )(

dttgs )(

2

0

4 62

1

ttv 120 msv

Displacement

Velocity

Acceleration

DifferentiateIn

tegr

ate

A particle moves in a straight line. Its velocity t s after leaving a fixed point on the line is v ms-1 where v = t + 0.1t2. Find an expression for the acceleration of the particle at time t and the distance travelled by the particle from time t = 0 until the instant when its acceleration is 2.8ms-2.

Puzzle TimeSome solutions

require double diff. or integration

Calculus Techniques Required

• You will be required to use ideas from C3 also.

atat aeet

)(d

d

ataatt

cos)(sind

d

ataatt

sin)(cosd

d

cea

te atat 1

d

cata

tat cos1

dsin

cata

tat sin1

dcos

Independent Study

Exercise A p 46 (solutions p 153)