As h 21b Onthemove

8/12/2019 As h 21b Onthemove

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Distance (x) and Displacement (s)

Distance (x) the length of the path moved by an object

scalar quantity

SI unit: metre (m)

Displacement (s)

the length and direction of the straight linedrawn from objects initial position to its final

position vector quantity

SI unit: metre (m)


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Speed (v)

average speed = distance change

time taken

vav= x / t

scalar quantity

SI unit: ms -1

Instantaneous speed (v) is the rate ofchange of distance with time: v = dx / d t


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Velocity (v)

average velocity = displacement change

time taken

vav= s / t

vector quantity

direction: same as the displacement change

SI unit: ms -1

Instantaneous velocity (v) is the rate of change of

displacement with time: v = ds / dt


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Speed and Velocity Conversions

1 kilometre per hour (km h-1)= 1000 m h-1

= 1000 / 3600 ms-1

1 km h-1= 0.28 ms-1

and 1 ms-1= 3.6 km h-1

Also:

100 km h-1= 28 ms-1= approx 63 m.p.h


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Complete

distance time speed

60 m 3 s 20ms-1

1400m 35 s 40 ms-1

300 m 0.20s 1500 ms-1

80 km 2 h 40km h-1

150 x 10 6km 8min 20s 3.0 x 108ms-1

1 km 3.03s 330 ms-1


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Speed and Velocity Question

Two cars (A and B)travel from Chertseyto Weybridge by theroutes shown

opposite. If both carstake 30 minutes tocomplete their

journeys calculate

their individualaverage speeds andvelocities.

car A: distance = 6km

car B: distance = 4km

displacement = 2km EASTChertsey

Weybridge


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Acceleration (a)

average acceleration = velocity changetime taken

aav= v / t

vector quantitydirection: same as the velocity change

SI unit: ms -2

Instantaneous acceleration (a) is the rate ofchange of velocity with time: a = dv / dt


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Notes:

1. Change in velocity:

= final velocity (v)initial velocity (u)

so: aav= (v u) / t

2. Uniform acceleration:This is where the acceleration remains

constant over a period of time.

3. Deceleration:This is where the magnitudeof the velocity

is decreasing with time.


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Acceleration due to gravity (g)

An example of uniform acceleration.

In equations a is substituted by g

On average at sea level:g= 9.81 ms-2downwards

gis often approximated to 10 ms -2YOU ARE EXPECTED TO USE 9.81 INEXAMINATIONS!!


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Question

Calculate the average acceleration of a car

that moves from rest (0 ms-1) to 30 ms-1over

a time of 8 seconds.

aav= (v u) / t

= (300) / 8

average acceleration = 3.75 ms-2


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Complete

Velocity / ms-1 Time

/ s

Acceleration

/ ms-2Initial Final

0 45 15 3

0 24 3 8

30 90 10 6

20 5 3 - 5

0 - 60 20 - 3


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Distance-time graphs

The gradientof a distance-time graph isequal to the speed


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Displacement-time graphs

The gradientof adisplacement-time graph is

equal to the velocity

The graph opposite showshow the displacement of an

object thrown upwards variesin time.

Note how the gradient fallsfrom a high positive value tozero (at maximum height) to alarge negative value.

Estimate the initial velocity of the object.

Initial gradient = (50)m / (0.50)s = 10 ms-1

Initial velocity = 10 ms-1


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Question

Describe the motion shown by the displacement-time graph below:

s/ m

t/ sA

B

C D

E


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Velocity-time graphs

With velocity-time graphs:

gradient

= acceleration

a = (vu) / t

The area under the curve

= displacement

s = [u x t] + [ (vu) x t]

velocity


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Question 1Describe the motion shown by the velocity-time graph below:

v/ ms-1

t/ s

AB

C

D E

F

v1

v2


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Question 2The graph shows the velocity-

time graph of a car. Calculateor state:

(a) the acceleration of the carduring the first 4 seconds.

(b) the displacement of the carafter 6 seconds.

(c) time T.

(d) the displacement after 11seconds.

(e) the average velocity of thecar over 11 seconds.

v/ ms-1

t/ sT

12

-10

4 6 11


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Question 3

Sketch the displacement-time graph for the car

of question 2.


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Question 4Sketch displacement and velocity time graphs for

a bouncing ball.Take the initial displacement of the ball to be hattime t = 0.

Use the same time axis for both curves and show

at least three bounces.


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The equations of uniform

acceleration

v = u + at

v2= u2+ 2as

s = (u + v) ts = u t + at2

v= FINAL velocity

u= INITIAL velocity

a= acceleration

t = time for the velocitychange

s= displacement during

the velocity change

THESE EQUATIONS ONLY APPLY

WHEN THE ACCELERATION

REMAINS CONSTANT


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Question 1

Calculate the final velocity of a car thataccelerates at 2ms -2from an initial velocity

of 3ms -1for 5 seconds.

v = u + atv= 3 + (2 x 5)

= 3 + 10

final velocity = 13 ms-1


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Question 2

Calculate the stopping distance of a car thatis decelerated at 2.5 ms -2from an initial

velocity of 20 ms -1.

v2= u2+ 2as0 = 202+ (2 x - 2.5 x s)

0 = 400 + - 5s

- 400 = - 5s- 400 / - 5 = s

stopping distance = 80 m


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Question 3

A stone is dropped from the edge of a cliff. Ifit accelerates downwards at 9.81 ms -2and

reaches the bottom after 1.5s calculate the

height of the cliff.s = ut + at2

s= (0 x 1.5) + (9.81 x (1.5)2)

s= (9.81 x 2.25)cliff height = 11.0 m


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Question 4

Calculate the time taken for a car toaccelerate uniformly from 5 ms -1to 12 ms -1

over a distance of 30m.

s = (u + v) t30 = (5 + 12) x t

30 = 8.5 x t

30 8.5 = ttime = 3.53 s


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Question 5

A ball is thrown upwards against gravity withan initial speed of 8 ms -1. What is the

maximum height reached by the ball?

v2= u2+ 2aswhere:

s= height upwards

u= 8 ms -1upwardsv= 0 ms -1(at maximum height)

a= - 9.81 ms -2(acceleration is downwards)


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Calculate the ?quantities

u/ ms-1 v/ ms-1 a/ ms-2 t/ s s/ m

2 14 0.75 ?

0 0.4 15 ?

16 0 - 8 ?

4 6 ? 20


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Calculate the other quantities

u/ ms-1 v/ ms-1 a/ ms-2 t/ s s/ m

2 14 0.75 128

0 6 0.4 15

16 0 - 8 2

4 6 0.5 20


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Projectile motion

This is where a body is moving in twodimensions. For example a stone beingthrown across a stretch of water has both

horizontal and vertical motion.

The motion of the body in two such

mutually perpendicular directions can betreated independently.


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Example 1

A stone is thrownhorizontally at a speed of8.0 ms-1from the top of avertical cliff.

If the stone falls vertically by

30m calculate the timetaken for the stone to reachthe bottom of the cliff andthe horizontal distancetravelled by the stone

(called the range).Neglect the effect of airresistance.

path of

stone

range

height

of fall


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Example 1

Stage 1

Consider vertical motion only

s = ut + at2

30 = (0 x t) + (9.81 x (t)2)

30 = (9.81 x (t)2)

30 = 4.905 x t2

t2 = 6.116

time of fall = 2.47 s

path of

stone

range

height

of fall


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Example 1

Stage 2Consider horizontal motion only

During the time 2.47 seconds thestone moves horizontally at a

constant speed of 8.0 ms-1

speed = distance / time

becomes:

distance = speed x time

= 8.0 x 2.47

= 19.8

range = 19.8 m

path of

stone

range

height

of fall


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Further Questions

(a) Repeat this example thistime for a cliff of height 40mwith a stone thrownhorizontally at 20 ms-1.

time of fall = 2.83 s

range = 56.6 m

(b) How would these valuesbe changed if air resistancewas significant?

time of fall - longerrange - smaller

path of

stone

range

height

of fall


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Example 2

A shell is fired at 200 ms-1

at an angle of 30 degrees to thehorizontal. Neglecting air resistance calculate:

(a) the maximum height reached by the shell

(b) the time of flight

(c) the range path of

shell

range

maximum

height30


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Example 2Stage 1 - Part (a)

Consider vertical motion only

At the maximum height, s

The final VERTICAL velocity, v= 0.

v2= u2+ 2as

0 = (200 sin 30)2+ (2 x - 9.81 x s) [upwards +ve]0 = (200 x 0.500)2+ (-19.62 x s)

0 = (100)2+ (-19.62 x s)

0 = 10000 - 19.62s

- 10000 = - 19.62ss= 10000 / 19.62

s= 509.7

maximum height = 510 m


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Example 2Stage 2Part (b)

Consider vertical motion onlyv = u + at0 = (200 sin 30) + (- 9.81 x t)0 = 100 - 9.81t-100 = - 9.81t

t= 100 / 9.81t= 10.19Time to reach maximum height = 10.19 s

If air resistance can be neglected then this is also the time

for the shell to fall to the ground again.Hence time of flight = 2 x 10.19

time of flight = 20.4 seconds


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Example 2Stage 3Part (c)

Consider horizontal motion only

During the time 20.38 seconds the shell moves horizontallyat a constant speed of (200 cos 30) ms-1

speed = distance / time

becomes:distance = speed x time

= (200 cos 30) x 20.38

= (200 x 0.8660) x 20.38

= 173.2 x 20.38

= 3530

range = 3530 m (3.53 km)


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Question

Repeat example 2 this time for a firing angle of 45.sin 45= 0.7071; 200 x sin 45= 141.4

maximum height = 1020 mtime of flight = 28.8 srange = 4072 m (4.07 km)

Note: 45yields the maximum range in this situation.path of

shell

range

maximum

height45

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