equations of motion - vcephysics.com

VCE Physics.comEquations of motion -

Equations of motion

1

• Displacement, velocity & acceleration

• Velocity-time graph

• The five kinematics equations

• Solving kinematics problems

• Gravity


• Displacement is a vector quantity - it is the change of position of an object.

• Velocity is a vector quantity - it is the rate of change of displacement with time.

• Acceleration is the rate of change of velocity with time. • There are five kinematics rules: each one requires three of the variables

to be known.

• These rules are based on constant acceleration.

• These rules apply to motion in one dimension (forward & back, up & down....). Directions can defined as positive or negative.

Displacement, velocity & acceleration

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x = displacement (m)t = time (s)

a = acceleration (m/s2)u = initial velocity (m/s)v = final velocity (m/s)


Velocity-time graphs

3

• The gradient of the graph gives the acceleration.

• The area under the graph gives the displacement.

Velocity (m/s)

Time (s)

u

v

t

Acceleration

Displacement1 m/s x 1 s = 1m


Defining the equations (1)

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• The gradient of the graph gives the acceleration.

Velocity (m/s)

Time (s)

u

v

t

a=

v-ut

v - u

t



5


Velocity (m/s)

Time (s)

u

v

x =

12

(u +v )tu

v

t

Trapezium area

=12

(u +v )t



6


Velocity (m/s)

Time (s)

u

v

t

v - u = at

ut

12

at2

Triangle area

=12

t × at

=12

at2

x =ut+

12

at2



7


Velocity (m/s)

Time (s)

u

v

t

12

at2

x = vt-

12

at2

vt



8

• We can also define a rule that is not dependent on time.

x =

12

(u +v )t

a=

v-ut

t =v-u

a

x =

(u +v )2

(v −u)a

x =

v 2 −u2

2a


• A car accelerates from the lights, 0 to 60 km/h in 4.0 seconds

• Find the distance covered in that time.

Solving kinematics problems (1)

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t = 4.0 s

Known information:

u = 0 m/s

v =

60 km/h3.6

= 17 m/s

x =

12

(u +v )t

x =

12

(0 + 17 m/s) × 4 s

x =34 m

(This is an average speed of 8.5 m/s over the 4 seconds.)


• The brakes & tyres of a car can provide a maximum deceleration of around 6 m/s2. If a car is traveling at 100 km/h, what is the minimum distance in which it could stop?


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a = -6 m/s2

Known information:

v = 0 m/s

u =

100 km/h3.6

= 28 m/s

x =65 m

x =

v2 - u2

2a

x = (0 m/s)2 - (28 m/s)2

2 × -6 m/s2

(Stopping distance varies with the square of speed.

2 x speed = 4 x stopping distance.)


• A tennis ball is hit upwards at a speed of 15 m/s.

• How high does the ball go? How long does it take to hit the ground?


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a = -10 m/s2

Known information:

v = -15 m/s

u = 15 m/s

If the ball lands at the height that it was hit from:

a=

v-ut

t =

- 15 m/s - 15 m/s-10 m/s2

t =

v-ua

t =3 sMaximum height at 1.5 s, v = 0

x =

12

(u +v )t

Acceleration due to gravityg ≈10 m/s2

x =

12

(15 m/s + 0 m/s)(1.5 s)

x = 11.25 m


• Objects falling (or rising up) experience an acceleration due to the force of gravity.

• g (on Earth) = 9.81 m/s2 ≈10 m/s2

• (Presuming that air resistance is insignificant: the projectile is heavy, small & not moving very fast.)

Gravity

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v = at

Time (s) Speed Distance

0 0 m/s 0 m

1 10 m/s 5 m

2 20 m/s 20 m

3 30 m/s 45 m

4 40 m/s 80 m

5 50 m/s 125 m

x =

12

at2

u = 0 m / s

0 m5 m

20 m

45 m

80 m

125 m


Graphs of motion under gravity

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v = u + at

x =ut +

12

at2

u = 0 m / s

a = -10 m/s2

v = −10t

a = constant

Velocity

time (s)

Acceleration

time (s)

x = −5t2

equations of motion - vcephysics.com

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