biology

CHAPTER 2

1D KINEMATICSLinear Motion

Kinematics definitions

• Kinematics – branch of physics; study of motion

• Position (x) – where you are located• Distance (d ) – how far you have traveled,

regardless of direction • Displacement (x) – where you are in

relation to where you started

Kinematics of Linear motion

• Defined as the studies of motion of an objects without considering the effects that produce the motion.

• There are two types of motion:– Linear or straight line motion (1-D)

• with constant (uniform) velocity• with constant (uniform) acceleration, e.g. free fall motion

– Projectile motion (2-D)• x-component (horizontal)• y-component (vertical)

3

Linear motion (1-D)

Distance, d

• scalar quantity.• is defined as the length of actual path between

two points.• For example :

– The length of the path from P to Q is 25 cm. 4

P

Q

s

5

vector quantity. is defined as the distance between initial point and final point in

a straight line. The S.I. unit of displacement is metre (m).

Example 2.1 :An object P moves 30 m to the east after that 15 m to the south

and finally moves 40 m to west. Determine the displacement of P

relative to the original position.

Solution :

Displacement,

N

EW

S

O

P

30 m

15 m

10 m 30 m

Distance vs. Displacement• You drive the path, and your odometer goes

up by 8 miles (your distance).• Your displacement is the shorter directed

distance from start to stop (green arrow).• What if you drove in a circle?

start

stop

The magnitude of the displacement is given by

and its direction is

2.1.3 Speed, v

• is defined as the rate of change of distance.• scalar quantity.• Equation: interval time

distance of changespeed

7

Δt

Δdv

m 181015 22 OP

south west tofrom 5610

15tan 1

θ

Speed, Velocity, & Acceleration

• Speed (v) – how fast you go

• Velocity (v) – how fast and which way; the rate at which position changes

• Average speed ( v ) – distance / time

• Acceleration (a) – how fast you speed up, slow down, or change direction; the rate at which velocity changes

v

9

interval time

ntdisplaceme of changeavv

12

12av tt

ssv

is a vector quantity. The S.I. unit for velocity is m s-1.

Average velocity, vav

is defined as the rate of change of displacement. Equation:

Its direction is in the same direction of the change in displacement.

Velocity,

Δt

Δsvav

constantdt

ds

10

t

s

0tv

limit

Instantaneous velocity, v is defined as the instantaneous rate of change of displacement. Equation:

An object moves in a uniform velocity when

and the instantaneous velocity equals to the average velocity at any time.

dt

dsv

Speed vs. Velocity

• Speed is a scalar (how fast something is moving regardless of its direction). Ex: v = 20 mph

• Speed is the magnitude of velocity.• Velocity is a combination of speed and direction. Ex:

v = 20 mph at 15 south of west

• The symbol for speed is v.• The symbol for velocity is type written in bold: v or

hand written with an arrow: v

Speed vs. Velocity

• During your 8 mi. trip, which took 15 min., your speedometer displays your instantaneous speed, which varies throughout the trip.

• Your average speed is 32 mi/hr.• Your average velocity is 32 mi/hr in a SE direction.• At any point in time, your velocity vector points

tangent to your path. • The faster you go, the longer your velocity vector.

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Therefore

Q

s

t0

s1

t1

The gradient of the tangent to the curve at point Q

= the instantaneous velocity at time, t = t1

Gradient of s-t graph = velocity

a

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interval time

velocityof changeava

vector quantity. The S.I. unit for acceleration is m s-2.

Average acceleration, aav

is defined as the rate of change of velocity. Equation:

Its direction is in the same direction of motion. The acceleration of an object is uniform when the magnitude of

velocity changes at a constant rate and along fixed direction.

Acceleration,

12

12av tt

vva

Δt

Δvaav

AccelerationAcceleration – how fast you speed up, slow

down, or change direction; it’s the rate at which velocity changes. Two examples:

t (s) v (mph)

0 55

1 57

2 59

3 61

t (s) v (m/s)

0 34

1 31

2 28

3 25

a = +2 mph / s a = -3 m / ss = -3 m / s

2

constantdt

dv

16

t

v

0ta

limit

Instantaneous acceleration, a is defined as the instantaneous rate of change of velocity. Equation:

An object moves in a uniform acceleration when

and the instantaneous acceleration equals to the average acceleration at any time.

2

2

dt

sd

dt

dva

Velocity & Acceleration Sign Chart

V E L O C I T Y ACCELERATION

+ -

+ Moving forward;

Speeding up

Moving backward;

Slowing down

- Moving forward;

Slowing down

Moving backward;

Speeding up

Example

Acceleration due to Gravity

9.81 m/s2

Near the surface of the Earth, all objects accelerate at the same rate (ignoring air resistance).

a = -g = -9.81 m/s2

Interpretation: Velocity decreases by 9.81 m/s each second, meaning velocity is becoming less positive or more negative. Less positive means slowing down while going up. More negative means speeding up while going down.

This acceleration vector is the same on the way up, at the top, and on the way down!

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Deceleration, a is a negative acceleration. The object is slowing down meaning the speed of the object

decreases with time.

Therefore

v

t

Q

0

v1

t1

The gradient of the tangent to the curve at point Q

= the instantaneous acceleration at time, t = t1

Gradient of v-t graph = acceleration

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Displacement against time graph (s-t)Graphical methods

s

t0

s

t0(a) Uniform velocity (b) The velocity increases with time

Gradient = constant

Gradient increases with time

(c)

s

t0

Q

RP

The direction of velocity is changing.

Gradient at point R is negative.

Gradient at point Q is zero.

The velocity is zero.

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Velocity versus time graph (v-t)

The gradient at point A is positive – a > 0(speeding up) The gradient at point B is zero – a= 0 The gradient at point C is negative – a < 0(slowing down)

t1 t2

v

t0 (a) t2t1

v

t0(b) t1 t2

v

t0(c)

Uniform velocity

Uniform acceleration

Area under the v-t graph = displacement

BC

A

dt

dsv

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From the equation of instantaneous velocity,

Therefore

vdtds

2

1

t

tvdts

graph under the area dedsha tvs

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A toy train moves slowly along a straight track according to the

displacement, s against time, t graph in figure below.

a. Explain qualitatively the motion of the toy train.

b. Sketch a velocity (cm s-1) against time (s) graph.

c. Determine the average velocity for the whole journey.

d. Calculate the instantaneous velocity at t = 12 s.

e. Determine the distance travelled by the toy train.

Example :

0 2 4 6 8 10 12 14 t (s)

2

4

6

8

10

s (cm)

25

Solution :

a. 0 to 6 s : The train moves at a constant velocity of

6 to 10 s : The train stops.

10 to 14 s : The train moves in the same direction at a constant velocity of

b.

0 2 4 6 8 10 12 14 t (s)

0.68

1.50

v (cm s1)

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Solution :

c.

d.

e. The distance travelled by the toy train is 10 cm.

12

12

tt

ssvav

s 14 tos 10 from velocity averagev

12

12

tt

ssv

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A velocity-time (v-t) graph in figure below shows the motion of a lift.

a. Describe qualitatively the motion of the lift.

b. Sketch a graph of acceleration (m s2) against time (s).

c. Determine the total distance travelled by the lift and its

displacement.

d. Calculate the average acceleration between 20 s to 40 s.

Example :

05 10 15 20 25 30 35 t (s)

-4

-2

2

4

v (m s1)

40 45 50

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Solution :

a. 0 to 5 s : Lift moves upward from rest with a constant acceleration of

5 to 15 s : The velocity of the lift increases from 2 m s1 to 4 m s1 but the acceleration decreasing to

15 to 20 s : Lift moving with constant velocity of

20 to 25 s : Lift decelerates at a constant rate of

25 to 30 s : Lift at rest or stationary.

30 to 35 s : Lift moves downward with a constant acceleration of

35 to 40 s : Lift moving downward with constant velocity

of

40 to 50 s : Lift decelerates at a constant rate of and comes to rest.

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Solution :

b.

t (s)5 10 15 20 25 30 35 40 45 500

-0.4

-0.2

0.2

0.6

a (m s2)

-0.6

-0.8

0.8

0.4

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Solution :

c. i.

05 10 15 20 25 30 35 t (s)

-4

-2

2

4

v (m s1)

40 45 50

A1

A2 A3

A4 A5

v-t ofgraph under the area distance Total 54321 AAAAA

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Solution :

c. ii.

d.

v-t ofgraph under the areant Displaceme

54321 AAAAA

12

12

tt

vvaav

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THE END…