Block 2

Distance, Speed andAcceleration

Displacement, Velocity and Acceleration

Special Rates of Change

What is to be learned?

• How rates of change apply to displacement, velocity and time.

Displacement

Distance (in a certain direction)

Velocity

Change in displacement over a period of time

Acceleration

Change in velocity over a period of time

x

v = dx/dt

a = dv/dt

(some call it speed!)

Original Formula

Derivative of Displacement

Derivative of Velocity

Ex An object is moving along x axis (cm) at time t(secs) according to equation:

x = 4t2 + 10t – t3

Calculate:

a) Displacement, Velocity and Acceleration

after 1 second and 3 seconds

b) Displacement and Velocity after

5 seconds

x = 4t2 + 10t – t3

displacement

t = 1

x = 4(1)2 + 10(1) – 13

= 13cm

velocity

v = dx/dt = 8t + 10 – 3t2

t = 1

v = 8(1) + 10 – 3(1)2

= 15 cm/sec

Original Formula

Derivative of Displacement

x = 4t2 + 10t – t3

velocity

v = dx/dt = 8t + 10 – 3t2

t = 1 v = 8(1) + 10 – 3(1)2

= 15 cm/sec

acceleration

a = dv/dt = 8 – 6t

t = 1 a = 8 – 6(1)

= 2 cm/sec/sec

cm/sec2

Derivative of Velocity

x = 4t2 + 10t – t3

displacement

t = 3 x = 4(3)2 + 10(3) – 33

= 39cm

velocity

v = dx/dt = 8t + 10 – 3t2

t = 3 v = 8(3) + 10 – 3(3)2

= 7 cm/sec

x = 4t2 + 10t – t3

velocity

v = dx/dt = 8t + 10 – 3t2

t = 3 v = 8(3) + 10 – 3(3)2

= 7 cm/sec

acceleration

a = dv/dt = 8 – 6t

t = 3 a = 8 – 6(3)

= -10 cm/sec2

x = 4t2 + 10t – t3

velocity

v = dx/dt = 8t + 10 – 3t2

t = 3 v = 8(3) + 10 – 3(3)2

= 7 cm/sec

acceleration

a = dv/dt = 8 – 6t

t = 3 a = 8 – 6(3)

= -10 cm/sec2

It is decelerating

x = 4t2 + 10t – t3

displacement

t = 5 x = 4(5)2 + 10(5) – 53

= 25cm

velocity

v = dx/dt = 8t + 10 – 3t2

t = 5 v = 8(5) + 10 – 3(5)2

= -25 cm/sec

x = 4t2 + 10t – t3

displacement

t = 5 x = 4(5)2 + 10(5) – 53

= 25cm

velocity

v = dx/dt = 8t + 10 – 3t2

t = 5 v = 8(5) + 10 – 3(5)2

= -25 cm/sec

It has changed direction

Displacement

Distance (in a certain direction)

Velocity

Change in displacement over a period of time

Acceleration

Change in velocity over a period of time

x (or h)

v = dx/dt (0r dh/dt)

a = dv/dt

Motion and Derivatives

(some call it speed!)

Original Formula

Derivative of Displacement

Derivative of Velocity

Ex Belinda throws a ball into the air

Its height (h m) after t secs is:

h = 4t – t2

Find

a) Its height, velocity and acceleration after 1 sec

b) What is its height when the velocity is zero?

h = 4t – t2

height (displacement)

t = 1 h = 4(1) – 12

= 3m

velocity

v = dh/dt = 4 – 2t

t = 1 v = 4 – 2(1)

= 2m/sec

h = 4t – t2

velocity

v = dh/dt = 4 – 2t

t = 1 v = 4 – 2(1)

= 2m/sec

acceleration

a = dv/dt = -2

t = 1 (or whatever!) a = -2 m/sec2

decelerating by 2m/sec2

b) Velocity zero?v = dh/dt = 4 – 2t

4 – 2t = 0 4 = 2t

t = 2velocity is 0 m/sec after 2 seconds

height? h = 4t – t2

t = 2 h = 4(2) – 22

= 4 m

Graphical explanation

h = 4t – t2

t

h

1

3

v = 0

4

2

The displacement (x cm) of a point after t secs can be

measured by the formula x = 10 – 6t + t2.

Calculate when the velocity will reach 10 cm/sec and

what the displacement will be after this time?

v = dx/dt = -6 + 2t

→ -6 + 2t = 10

→ 2t = 16

→ t = 8 secs

t = 8, x = 10 – 6(8) + 82

= 26 cm

Tricky Key Question