acceleration when an unbalanced force acts on an object it accelerates. an unbalanced force acting...

Acceleration

When an unbalanced force acts on an object it accelerates.

An unbalanced force acting on a stationary object will make it move.

An unbalanced force can make a moving object get faster or slower or change direction.

Acceleration is the change in velocity per unit time.

The acceleration or deceleration of an object is calculated using:

tu-v

atime (s)

acceleration

(ms-2)

initial velocity (ms-

1)

final velocity (ms-1)

Measuring Acceleration

DiagramElectronic

Timer

Light GateLength of Card

Trolley

Start Line

Finish Line

Stopwatch

Method

A trolley is fitted with a length of card. The length of card is measured using a ruler.

The trolley is released from rest, so the initial velocity is 0 ms-

1.

The time taken for the trolley to travel from the start to the finish line is measured using a stopwatch.

An electronic timer measures the time taken for the length of card to pass through the light beam. The timer starts when the card breaks the light beam and stops when the card leaves the light beam.

The final velocity is calculated using:

These measurements are used to calculate acceleration in the relationship shown:

timecard of length

v

tuv

a

Results

-1ms 0u

1-ms time

card of lengthv

s t

?a

Motion Graphs

There are three types of motion graph we will study:

1. A velocity - time graph.

2. A displacement – time graph.

3. An acceleration – time graph.

displacement-time

velocity-time acceleration-time

Constant Velocity

Constant Acceleration

t

s

t

v

t

a

t

s

t

v

t

a

Constant Deceleration

t

a

t

v

t

s

Drawing Graphs

Each velocity – time graph has a corresponding acceleration – time graph and displacement- time graph.

Example

The following velocity – time graph describes a journey.

time / s

velocity / ms-1

3 10 12

12

0

Draw the corresponding acceleration – time graph.

tuv

a

3012

2ms 4a

tuv

a

71212

2ms 0a

tuv

a

2120

2ms -6a

0-3 seconds 3-10 seconds

10-12 seconds

time / s

acceleration / ms-2

3

4

010 12

-6

Q1. Copy out the following velocity - time graph and underneath it draw the corresponding acceleration – time graph.

time / s

velocity / ms-1

4 13 16

20

0 9

24

tuv

a

4020

2ms 5a

2ms 0a t

uva

42024

2ms 1a

0-4 seconds 4-9 seconds 9-13 seconds

tuv

a

3240

2ms 8a

13-16 seconds

acceleration / ms-2

time / s4

5

09 13

- 8

16

1

Q2. Using the following acceleration – time graph of an object starting from rest, draw the corresponding velocity – time graph.

acceleration / ms-2

0

5

-24 8 1

8time / s

atuv

450 -1ms 20v

atuv

102-20

2020 -1ms 0v

velocity / ms-1

0

20

4 8 18

time / s

Q3. Copy out the following velocity – time graph and underneath it draw the corresponding acceleration – time graph (after appropriate calculations).

velocity / ms-

1

time / s

3

15

0

-9

3 7 9 14 2418

tuv

a

3315

2ms 4a

0-3 seconds 3-7 seconds

2ms 0a t

uva

2153

2ms -6a

7-9 seconds

9-14 seconds

2ms 0a t

uva

439-

2ms -3a

14-18 seconds

tuv

a

6

9-0

2ms 1.5a

18-24 seconds

acceleration / ms-2

0

4

-6

3 7 9 14 2418 time / s

Motion and Direction

In the velocity-time graphs so far, motion has been in only one direction.

A velocity time graph however can show two different directions; one direction is positive, while the other is negative.

Example 1

A car travelling along a straight section of road has the following velocity-time graph.

(a) Calculate the distance travelled by the car.

(b) Calculate the displacement of the car.

forward motion

backward motion

(a) Distance travelled is the area under velocity-time graph.

2010021

d1 105021

d2

10100d3 105021

d4

m 1,000

m 1,000 m 250

m 250

(b) Displacement is the forwards motion, less the backward motion.

2501,000250 - 1,000 ntdisplaceme

m 500

m 2,500dtot

Worksheet – Graphs of Motion

Q1 – Q9

Ball Thrown Into Air

A ball is thrown directly upwards into the air.

It rises into the air and falls back down to the thrower.

The velocity – time graph and corresponding acceleration – time graph are shown.

velocity / ms-1

time / s

acceleration / ms-2

time / s

-9.8

falling

Bouncing Ball (No Energy Loss)

A ball is dropped from a height to the ground.

The ball bounces twice with no energy loss and is then caught.

The velocity - time and acceleration - time graphs are as follows:

time / s

velocity / ms-1

downwards (falling): - ve

upwards (rising): + ve

rising

falling

rising

hits ground

max height

time / s

acceleration / ms-2

-9.8

rising

TASK

Re-draw the velocity-time graph for the bouncing ball, taking downward motion as positive, and upward motion as negative.

falli

ng

time / s

velocity / ms-1

downwards (falling): + ve

upwards (rising): - ve

hits ground

max height

rising

falli

ng

Bouncing Ball (With Energy Loss)

A ball is dropped from a height to the ground.

Kinetic energy is lost with each bounce.

The velocity - time and acceleration - time graphs are as follows:

time / s

velocity / ms-1

time / s

acceleration / ms-2

-9.8

TASK

Re-draw the velocity-time graph for the bouncing ball, taking downward motion as positive, and upward motion as negative.

Worksheet – More Graphs of Motion

Q1 – Q7

Equations of Motion

1st Equation of Motion

tuv

a

t auv

t auv

There are three equations of motion.

You must be able to use and derive these three equations of motion.

2nd Equation of Motion

To derive the second equation of motion, the velocity-time graph shown is used as a starting point.

time/s

velocity/ms-

1

u

v

t

} v - u

Displacement (s) is the area under a velocity time graph.

area areas

u-v t21

ut

at t21

ut s

but from equation (1):

atuv

atu-v

so we can rewrite as:

2at21

ut s

atuv

2atuv2

atuatuv2 2222 ta2uatuv

taking a common factor of 2a gives

222 at

21

ut2auv

and since s = ut + ½at2

2asuv 22

3rd Equation of Motion

This equation links final velocity (v) and displacement (s).

Direction

When using the equations of motion, it is essential that direction is considered.

In these examples, upward motion is taken as positive, so any downward motion is taken to be negative.

Example

A helicopter is travelling upwards with a velocity of 25 ms-1.

A package is released and hits the ground 14 s later. *

*

*

released

stationary ( 0 ms-1

)

hits ground

Path of Package

This example has motion in two directions.

It is necessary to distinguish between the two directions.

Choose the upward direction as positive!

(a) How long will it take the package to reach its maximum height? (2)

?t 1ms 25 u

1ms 0v 2ms 9.8a

atuv

t9.8-250

25t 9.8 s 2.55t

(b) How high as it climbed since being released?(2)

?s 1ms 25 u

s 2.55t 2ms 9.8 a

2atuts2

1

22.559.8-2.5525 2

1

31.8663.75

m 31.9s

(c) Calculate the velocity of the package just before it hits the ground.(2)

?v1ms 25 u

s 14 t 2ms 9.8a

atuv 149.8-25

137.2-251ms 112.2v

The negative indicates travelling downwards.

(d) How high above the ground is the helicopter when the package is released? (2)

?s 1ms 25 u

s 14t 2ms -9.8a

2atuts2

1

2149.8-21

1425

960.4350

m 610.4s

So the helicopter is 610.4 m above the ground.

Worksheet – Equations of Motion

Q1 – Q15

Acceleration Due To Gravity

Diagram

Ball Bearing

Trap Door

Timer

Method

The ball bearing is released from rest, so initial velocity (u) is 0 ms-1.

The displacement (s) of the ball bearing is the distance between the release point and trap door, and is measured using a metre stick.

An electronic timer measures the length of time taken for the ball bearing to reach the trap door.

-1ms 0um s

s t

?a

2atuts2

1

Calculation

Projectiles (Half Trajectory)

An object projected sideways through the air will follow a curved trajectory.

horizontal motion (steady speed)

vertical motion

The horizontal and vertical motions should be treated separately.

Time is the only quantity common to both.

accelerates downwards at -9.8 ms-

2

tD

V HH

vh

vv

This is an example of a ‘half-trajectory.’

GREEN – actual motion

RED – vertical motion

BLUE – horizontal motion

At any point in its trajectory, the velocity of a projectile has two components.

• one vertical, VV

• the other horizontal, VH

The resultant velocity is found drawing a vector diagram and add the vectors together, TIP to TAIL.

Vector Diagram

horizontal velocity

vertical velocity

resultant/actual

velocity

30 ms-1

Example

A ball is kicked horizontally off an embankment, with a velocity of 30 ms-1.

It lands 24 m from the base of the embankment.

(a) Calculate how long the ball was in flight.

24 m

tD

V HH

t24

30

3024

t

s 0.8t common to

horizontal and vertical motions

Horizontal Vertical

-1ms 30u

m 24s

-2ms -9.8a

(b) Calculate the horizontal velocity just before hitting the ground.

s 0.8t

s 0.8t

-2ms 0a

-1ms 0u

travels horizontally at steady speed

– no acceleration horizontally

not initially falling down, so speed of

zero in vertical direction

acted upon by gravity

Horizontal

atuv

0.8030 -1ms 30v

(c) Calculate the vertical velocity just before hitting the ground.

Vertical

atuv 0.89.8-0

-1ms -7.84v

(d) How high is the embankment?

Vertical

-2ms -9.8as 0.8t

-1ms 0u-1ms -7.84v

means 7.84 ms-1 downwards

2at21

uts

20.89.8-21

0.80

m 3.14s

so height of the embankment is 3.14 m

means ball fell through distance of

3.14 m

(e) Calculate the resultant velocity of the ball, just before hitting the ground.

30 ms-

1

-7.8 ms-1

velocity

θ

Size

By Pythagoras:

222 cba

222 7.84-30velocity resultant 61.5900

961.5velocity resultant -1ms 31

Direction

hyp

adjθ cos

3130

θ cos

0.97cosθ 1

14.6θ

horizon below 14.6 of angle at ms 31velocity resultant -1

30 ms-

1

-7.8 ms-1

velocity

θ

Q1. A ball is kicked off a cliff with a horizontal speed of 16 ms-

1.

The ball hits the ground 2.2 s later.

(a) Calculate the height of the cliff.

(b) Calculate the distance between the foot of the cliff and where the ball lands.

(c) Calculate the vertical component of the balls velocity just before it hits the ground.

(d) Calculate the balls velocity as it hits the ground.

23.7 m

35.2 m

21.6 ms-1

26.9 ms-1 at angle of 53.5° below

horizon

You may want to draw a diagram to help you get started !!!

Q2. A ball is kicked off a cliff with a horizontal speed of 22 ms-1. the ball hits the ground 1.5 s later.

(a) Calculate the height of the cliff.

(b) Calculate the horizontal distance from the foot of the cliff, to where the ball lands.

(c) Calculate the vertical component of the balls velocity as it hits the ground.

(d) Calculate the balls actual velocity as it hits the ground.

11 m

14.7 ms-1

26.5 ms-1 at angle of 34° below

horizon

33 m

You may want to draw a diagram to help you get started !!!

Does Projectile Theory Work?

Diagramball-bearing

h

d

sand

Measurements

Horizontal Velocity measure distance ball-bearing travels along desk and divide by time

taken

Vertical Displacement measure height of desk from floor

Calculation

Calculate the time of flight.m s

-2ms -9.8a-1ms 0u

?t

2at21

uts Vertical

Now calculate the horizontal displacement.

-1H ms v

Horizontal

s t

?s

tvs H

Experimentally

The horizontal displacement was measured experimentally using

a metre stick to be m.

Projectiles (Full Trajectory)

A projectile does not need to be an object falling, but could be an object fired at angle to the horizontal.

The subsequent motion would bemax height

θ

If air resistance is ignored, the trajectory has an axis of symmetry about the mid point (maximum height).

So the time taken to reach the maximum height is the same as the time taken to fall back to the ground.

Various calculations can be made, but firstly, the initial velocity must be split into its horizontal and vertical components.

Horizontal

a = 0 ms-2

Vertical

a = -9.8 ms-2

Example 1

A golf ball is hit off the tee at 48 ms-1 at angle of 20° to the horizontal.

Calculate the horizontal and vertical components of the initial velocity.

Horizontal

20°

48 ms-

1

hypadj

θ cos

48v

20 cos H

cos2048vH -1

H ms 45.1v

Vertical

hypopp

θ sin

48v

20 sin V

20 sin48vV -1

V ms 16.4v

48 ms-1

20°

VH

VV

Example 2

An arrow is projected into the air with a velocity of 38 ms-1 at an angle of 25° to the horizontal.

38 ms-1

250

(a) Calculate the horizontal and vertical components of the initial velocity.

Horizontal

hypadj

θ cos

38v

25 cos H

25 cos38vH -1

H ms 34.4v

Vertical

hypadj

θ sin

38v

25 sin V

25 sin38vV -1

V ms 16.1v

38 ms-1

25°

VH

VV

(b) Calculate the arrow’s maximum height.

Vertical

-2ms -9.8a-1ms 16.1u

-1ms 0v

?s

2asuv 22

s-9.8216.10 22

s 19.6259.210

259.21s 19.6

m 13.2s

(c) Calculate the time taken for the arrow to reach its maximum height.

Vertical

-2ms -9.8a-1ms 16.1u

-1ms 0vm 13.2s

?t

atuv

t9.8-16.10 16.1t 9.8

s 1.64t

(d) Calculate the total time of the arrows flight.

down time up timetime total

1.64 1.64

s 3.28

(e) Calculate the horizontal distance travelled by the arrow until impact with the ground.

Horizontal

-1ms 34.4u-2ms 0as 3.28t

?s

2at21

uts

23.28021

3.2834.4

112.8ms

(f) Calculate the arrow’s velocity 0.5 s after being fired.

Firstly calculate the vertical component of velocity (horizontal component is constant, since a = 0 ms-2)

Vertical

-2ms -9.8a-1ms 16.1u

s 0.5t ?v

atuv

0.59.8-16.1

4.9-16.11ms 11.2v

Now calculate the actual velocity after combining the vertical and horizontal components of the velocity after 0.5 s.

v

θ

34.4 ms-

1

11.2 ms-

1

Size

By Pythagoras: 222 cba

222 11.234.4velocity 125.441,183.36

1,308.8velocity-1ms 36.2

Direction

adjopp

θ tan

34.411.2

θ tan

0.326tanθ 1

18θ

Velocity of the arrow after 0.5 s is:

36.2 ms-1 at angle of 18° above the horizon

Q1. A shell is fired from a gun with a velocity of 72 ms-1 at an angle of 60° to the horizontal.

(a) Calculate the horizontal and vertical components of the initial velocity.

(b) Calculate the maximum height reached.

(c) Calculate the time taken for the shell to reach it’s maximum height.

(d) Calculate the total time of flight.

(e) Calculate the horizontal range of the shell.

(f) Calculate the shells velocity after 2.3 s.

VH = 36 ms-1

VV = 62.4 ms-1

199 m

6.4 s

12.8 s

458 m

53.7 ms-1 at angle of 48° above the horizon

Q2. An arrow is fired with a velocity of 50 ms-1 at an angle of 30° to the ground.

(a) Calculate the time taken for the arrow to reach its maximum height.

(b) Calculate the maximum height reached by the arrow.

(c) Calculate the time the arrow is in flight.

(d) Calculate how far away from the firing point the arrow will land.

(e) Calculate the actual velocity of the arrow 1s after it is fired.

2.55 s

31.89 m

5.1 s

5.1 s

45.89 ms-1 at angle of 19.3° above horizon