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Kinematics H1/H2

Assessment objectives

Candidates should be able to:

(a) define displacement, speed, velocity and acceleration.

(b) use graphical methods to represent displacement, speed, velocity and acceleration.

(c) find the distance traveled by calculating the area under a velocity-time graph.

(d) use the slope of a displacement-time graph to find the velocity.

(e) use the slope of a velocity-time graph to find the acceleration.

(f) derive, from the definitions of velocity and acceleration, equations which represent uniformly-accelerated motion in a straight line.

(g) use equations which represent uniformly-accelerated motion in a straight line, including falling in uniform gravitational field without air resistance.

(h) describe qualitatively the motion of bodies falling in a uniform gravitational field with air resistance.

(i) describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction.

References:

1 Advanced Level Physics – Nelkon & Parker2 A-Level Physics – Roger Muncaster

Prepared by:Mr Tan Kia YenJanuary 2007

Tampines Junior College

Kinematics H1/H2

Introduction

This is a branch of mechanics which deals with the description of the motion of objects, without references to the forces which act on the system.

In kinematics, we will study the following two types of motion:

(i) one dimensional motion, i.e. along a straight line(ii) two dimensional motion – projectile motion.

Displacement, Velocity and Acceleration

(a) Displacement

Displacement is the distance travelled along a specific direction.

It is a vector quantity.Its symbol is s.Its SI unit is the metre, m.

It differs from distance in that distance is the total length moved from a reference point. Displacement is the shortest distance between the initial and final position of the body.

(b) Velocity

Speed is the rate of change of distance with time.

It is a scalar quantity.Its SI units are m s-1

Average speed, <speed> =

Velocity is the rate of change of distance along a specific direction, or simply the rate of change of displacement.

It is a vector quantity.Its SI units are m s-1

Its symbol is u (initial speed) or v (final speed).

Average velocity, <v> =

Note: <speed> is not always equal to <velocity>.


Kinematics H1/H2

(c) Acceleration

Acceleration is the rate of change of velocity.

It is a vector quantity.Its SI units are m s-2

Its symbol is a.

Retardation or deceleration is a term used to describe a decrease in the magnitude of velocity with time. It occurs when the acceleration and the velocity vectors are in opposite directions or opposite signs.

A change in velocity can be(a) a change in magnitude e.g. body speeding up or slowing down.(b) a change in both magnitude and direction.(c) a change in direction only e.g. circular motion.

The acceleration of a body is constant or uniform if its velocity changes at a constant rate.

Uniformly accelerated motion

In many practical applications, the acceleration of an object is constant. E.g. a toy car sliding down a slope, a ball being thrown in the air. The velocity is changing at a steady rate. This type of motion is called uniformly accelerated motion.

Acceleration due to gravity, g

This type of acceleration is produced by the gravitational field of the earth and it is always directed downward relative to the earth’s surface. It can be taken to be constant at 9.81 m s-2 (unless otherwise stated) for all bodies near the earth’s surface (ignoring air resistance). This downward acceleration due to free fall is known as acceleration due to gravity.

Graphical Representation of Motion

(a) Displacement-Time Graph

From this graph, you can find useful information about the object’s

(i) instantaneous displacement (the displacement of a body at any instant of time)(ii) instantaneous velocity (gradient of the graph at a particular point)(iii) average velocity


Kinematics H1/H2

eg) Consider a car traveling along the x-axis. What deduction can be made about the car’s motion if its displacement-time relation is represented by (i) Graph 1 and (ii) Graph 2?

(i) From graph 1, we can deduce the following:

(a) The displacement x carries only positive values. This implies that the car moves along the positive x direction.

(b) The displacement x increases uniformly with time, thus the velocity of the car remains constant.

(c) Average velocity = <u>=

(ii) From graph two, we can deduce the following:

(a) Since the displacement x remains positive throughout the motion, the car’s positions remain to the right of the origin O.

(b) At points A and C, the car is stationary, since its displacement x is not changing at these two points.

(c) Around B, the displacement x increases uniformly with time, so the car must be traveling in the positive x-direction with uniform velocity.

(d) After point C, the displacement x of the car is decreasing. This implies that the car is traveling in the opposite direction, i.e. the negative direction.

(e) The gradient at D is greater than at E. Thus, the magnitude of the velocity at D is greater than at E.

(f) Around E, the displacement x decreases with time at a decreasing rate up to F. This shows that the car eventually stops at F.


Graph 1 Graph 2

Kinematics H1/H2

Examples of Displacement-Time Graphs

(b) Velocity-Time Graph

From this graph, you can find useful information about the object’s

(i) instantaneous velocity (the velocity of a body at any instant of time)(ii) instantaneous acceleration (gradient of the graph at a particular point)(iii) displacement (area under the graph)


s

t t

t t

s

s s

Motion under constant velocity Motion under constant deceleration

Motion of a particle being projected vertically upwards and then return to to point of projection

Motion under constant acceleration

Kinematics H1/H2

eg) The motion of a body moving along a straight line is given as shown in the graph below. What deductions can be made about the body’s motion?

Deductions:

A-B Body moves from rest with velocity increasing at a constant rate, or uniform acceleration.

B-C Body moving with velocity increasing at a decreasing rate.

C-D Body is moving with magnitude of velocity decreasing at an increasing rate, body is decelerating.

D-E Body moving with magnitude of velocity decreasing at a decreasing rate, body is still decelerating, but at a slower rate compared with C-D.

E-F Stationary.

F-G Body moving in opposite direction with magnitude increasing, or accelerating in opposite direction.

G-H Constant velocity.

H-I Body undergoing constant deceleration and comes to rest at I.

Total distance moved = area 1 + area 2Net displacement = area 1 – area 2


v

tA

B

C

D

EF

G H

IArea 1

Area 2

Kinematics H1/H2

eg) The figure below shows a velocity-time graph for a journey lasting 65 s. It has been divided up into six sections for each case of reference.

(a) Using information from the graph obtain

(i) the velocity 10 s after the start,(ii) the acceleration in section A.(iii) the acceleration in section E,(iv) the distance travelled in section B,(v) the distance travelled in section C.

(b) Sketch the corresponding displacement-time graph.

(a)(i) velocity 10 s after the start = 20 m s-1

(a)(ii) acceleration in Section A =

= 2.0 m s-2

(a)(iii) acceleration in Section E =

= 7.0 m s-2

(a)(iv) Distance travelled in Section B = 20 x 15= 300 m


10 20 30 40 50 60 70

10

20

30

-5 5

v/m s-1

t/s

A B C D E F

Kinematics H1/H2

(a)(v) Distance travelled in Section C = ½(20 + 30)(10)= 250 m

Examples of Velocity-Time Graphs

(c) Acceleration-Time Graph


v

t t

t t

v

v v

i : zero accelerationii : uniform accelerationiii : uniform deceleration

Decreasing acceleration

Object being thrown vertically upwards and then return to its point of projection.

One bounce of a ball

i

ii

iii

10 20 30 40 50 60 70

Distance from start

t/s

A B C D E F

Kinematics H1/H2

This gives important information about the instantaneous acceleration and the velocity after a time t.

Equations of motion for uniformly accelerated motion in a straight line.

In most practical situations, the acceleration of a body is constant. This motion is called uniformly accelerated motion.

Suppose that a body is moving with constant acceleration a and that in a time interval t, its velocity increases from u to v and its displacement increases from 0 to s.

since a =

hence v = = at + constant

-------------------(1)

Average velocity =

and displacement = (average velocity) x (time)

-------------------(2)

substituting (1) into (2), we have:

-----------------(3)

eliminating t and substituting (1) into (2), we have:

hence, ------------(4)

Note:These 4 equations are called kinematics equations or equations of motion for uniformly accelerated motion. They are only valid for cases of uniform acceleration in a straight line. Appropriate sign conventions must be used.

Sign Conventions


v = u + at

v2=u2 + 2as

Kinematics H1/H2

If the convention is set such that upwards is taken to be positive, then

(i) velocity of body moving down is negative(ii) displacement of body below the starting (reference) point is negative(iii) downward directed acceleration is negative

eg) A ball is projected upwards from point A. Identify the sign of the displacement, velocity and acceleration during its flight, taking the upwards direction as positive.

eg) A motorist travelling at 13 m s-1 approaches traffic lights which turn red when he is 25 m away from the stop line. His reaction time (time between seeing the red lights and applying the brakes) is 0.70 s and the condition of the road and his tyre is such that he cannot slow down at a rate of more than 4.5 m s-2. If he brakes fully, how far from the stop line will he stop, and on which side of it?

s1 = ut = 13 x 0.70 = 9.1 m

v2 = u2 + 2as 02 = (+132) + 2(-4.5)s2 s2 = 18.8 m

Total distance s = s1 + s2 = 9.1 + 18.8 = 27.9 m 25 m The car overshot the traffic light.

eg) A man throws a ball vertically upwards with a velocity of 20 m s-1. Neglecting air resistance, find(i) the maximum height reached(ii) the time taken for the ball to return to the man’s hand.

(i) At maximum height, v = 0 m s-1,

Using v2 = u2 + 2as


A

1

2

3

45

s v a

1 + + -g

2 + 0 -g

3 + - -g

4 0 - -g

5 - - -g

v

U

W

Fv (= kv)

When an object falls through a fluid, it experiences three forces: its weight W, an upthrust U and a viscous force Fv.

Kinematics H1/H2

02 = (+20)2 + 2(-9.81) ss = 20.4 m.

(ii) using v = u + at0 = (+20) + (-9.81)tt = 2.04 s

Total time of flight = 2 x 2.04 = 4.08 s

Effect of Air Resistance

So far in our discussion we have ignored the effect of air resistance. The fact is that when a body moves in a fluid, it will experience a drag force called the viscous force Fv. This resistive force is velocity dependent i.e. the faster the velocity, the greater the viscous force. The proportionality constant k is affected by the size and shape of the object and the viscosity of the fluid.

As an object falls through the air, it will be accelerated by gravitational pull W. Fv is small initially since the velocity is small.

However, as the velocity picks up, Fv becomes more and more significant, eventually large enough to balance W:

U + Fv = W

When this condition is reached, the object will stop accelerating and continues to fall at constant velocity from then onwards. This constant velocity is known as the terminal velocity.

eg) A commando launches his parachute 20 s after jumping off a helicopter. Draw a graph to illustrate how his velocity vary with time from the moment he jumps of his helicopter till the moement he lands safely.


v

20

A

B

C

Kinematics H1/H2

At A:As the commando falls through the air, he experiences a stronger and stronger viscous force. As a result, his acceleration gradually decreases from 9.8 m s-2 to zero. Before he opens his parachute, he is already moving at terminal velocity which has a large magnitude.

At B:After 20 s he opens he parachute. This increases his cross-sectional area immediately and he experiences a much larger viscous force. At this moment, U + Fv > W. So he starts to decelerates until U + Fv = W.

At C:When U + Fv = W is achieved, he reaches a new terminal velocity which is much lower than the previous one. This will allow him to land safely.

Projectile Motion

A projectile motion can be considered as a combination of two independent components of motion:

(i) horizontal motion with constant speed throughout, zero acceleration (since there is no acceleration horizontally), assuming there is no air resistance.

(ii) vertical motion with uniform acceleration (for example due to gravity g).


t

Kinematics H1/H2

Analysis of projectile motion

(a) Object projected horizontally from a height h above ground level

Taking downwards as positive:

Time of flight: vertically:

Using

h = 0(t) + ½ gt2

Range, R : horizontally:

Using

R = u( )

Velocity at any time t : horizontally: vx = uvertically: vy = gtresultant velocity,

tan =


u

hvx

vy

R

Kinematics H1/H2

(b) Object projected at an angle from ground level

Taking upwards as positive:

Maximum height: vertically:Using v2 = u2 + 2as

0 = (usin)2-2gh

h=

Time of flight: vertically:Using v = u + at

0= usin - gt

therefore, time of flight = 2t=

Range, R : horizontally:

Using

R = (ucos)( )=

R is maximum when = 45o.

Velocity at any time t : horizontally: vx = uvertically: vy = gtresultant velocity,

tan =

Position of object at any time t:

Vertically:

= -----------(1)

horizontally:-----------(2)

combining (1) and (2):


uh vy

R

vx

v

Kinematics H1/H2

which is of the form y = bx + cx2 .

This is a parabolic equation and so the trajectory (or path) of our projectile is a parabola.

eg) A bomber, flying at a horizontal speed of 360 km h-1 and at an altitude of 2 km above sea level, wishes to attack an enemy vessel. Calculate the angle of sight , at which the bomber should release its bomb so that it would most probably hit the vessel.

Vertically:

Using

-2000 = 0 + (-9.81) t2

t = 20.2 s

horizontally:

= 44.7o


2 km

360 km h-1

x

Kinematics H1/H2

eg) A stone is projected with an initial velocity of 12 m s-1 at angle of 30 above the horizontal from a cliff top which is 75 m above the sea level. Find

(a) the time taken for the stone to reach the sea, and

(b) the position of the stone from the cliff when it reaches the sea.

(a) Vertically:

Using

t = 4.5 s

(b) horizontally:

Using

s = ut = 12cos30o x 4.5 = 46.8 m


75 m

s

12 m s-1

30

Kinematics H1/H2

Tutorial:

Basic Questions:

1) What are the conditions that must be satisfied in order for the kinematics equations to be valid?

2) a) Can a body have zero velocity and still be accelerating?b) Can a body have a constant speed and still have a varying velocity?c) Can a body have a constant velocity and still have a varying speed?d) Can the direction of the velocity of a body change when its acceleration is constant?e) Can a body be increasing in speed as its acceleration decreases?f) Can a body have constant acceleration and variable velocity?g) Can a body have constant velocity and variable acceleration?h) Can the average velocity over any time interval differ from the instantaneous velocity

at any instant when the velocity is constant?

3) The following table shows the timing that a student took to complete each of the six laps of his 2.4 km run.

Lap 1 2 3 4 5 6

Timing/s 97 104 115 129 132 122

What was the student’s average speed for the run?

4) A body is under the influence of the earth’s gravitational field and subject to the acceleration g = 9.81 m s-2. Considering only the motion in the vertical direction, complete the following table:

velocity acceleration example of situation is the object getting faster, slower or neither?

+ +

- +

0 +

+ -

- -

0 -


Kinematics H1/H2

Application Questions:

5) A brick is dislodged from a tall building and falls vertically under gravity. Which one of the following curves represents the variation of its height h above the ground with time t if air resistance is negligible?

6) The velocity-time curves of two cars A and B are shown below:

Just as car A starts up, car B passes it, moving at a steady velocity. (a) How long does it take car A to be going as fast as car B?(b) At that time, by how far is car B ahead of car A?(c) Which car is ahead, and by how much, at the end of 40 s?(d) At what time does car A catch up with car B?

7) The figure below shows the velocity of a particle plotted as a function of time. The total distance travelled by the particle over the interval t = 0 s to t = 3.0 s is 30 m.

(a) What is the average velocity over the interval t = 0 s to t = 1.5 s?


t

h

tA

h

tB

h

tC

h

D

20

40

60

20 40 600

v / m s-1

t / s

Car A

Car B

t / s1 32

20

- 20

v / m s-1

Kinematics H1/H2

(b) What is the particle’s maximum displacement from its position at t = 0 s?(c) Sketch the particle’s displacement and acceleration as a function of time.

8) A cricketer throws a ball vertically upwards and catches it 3.0 s later. Neglecting air resistance, find

(a) the speed with which the ball leaves his hands,(b) the maximum height to which it rises.

Without carrying any calculations, explain how air resistance would affect

(c) the time taken for the ball to reach its maximum height,(d) the maximum height to which it rises.

9) A balloon is ascending at the rate of 12 m s -1 at a height of 80 m above the ground when a package is dropped. How long does it take for the package to reach the ground?

10) Give physical explanations as to why

(i) negative accelerations of cars are usually numerically larger than positive accelerations.

(ii) parts of acceleration-time graphs for cars can be almost vertical but velocity-time graphs for vehicles can never be vertical.

11) An aeroplane flying in a straight line at a constant height of 500 m with a speed of 200 m s-1 drops an object. The object takes a time t to reach the ground and travels a horizontal distance d in doing so. Ignoring air resistance, find the values of t and d.

12) A stone is thrown from the top of a cliff with a velocity of 15.0 m s-1 at an angle of 60o to the horizontal. The cliff is 42.0 m high. Calculate

(a) the maximum vertical height of the stone above its point of projection,(b) the velocity of the stone just before it hits the ground,(c) the horizontal distance covered by the stone when it reaches the ground.

13) A boy throws a stone in a direction inclined at angle tan-1 (2.4) to the horizontal, directly towards a building with a large flat roof. The near edge of the roof is at a horizontal distance 7.5 m from the boy and the roof is 4.0 m above the point of projection. If the stone passes vertically above the near edge of the roof 1.5 s after being thrown, find(a) the velocity of the projection(b) the height of the stone above the roof as it passes over its near edge(c) the distance from the near edge of the point where the stone hits the roof.(d) the magnitude and direction of the stone’s velocity when it hits the roof.

Challenging Questions:


Kinematics H1/H2

14) Two cyclists travel at a uniform speed of 10 km h-1 towards each other. At the moment when they are 20 km apart, a bumble bee flies from the front wheel of one of the bikes at a uniform speed of 25 km h-1 directly to the wheel of the other bike. It touches it and turns around in a negligible short time and returns at the same speed to the first bike, whereupon it touches the wheel and instantaneously turns around and repeats the back-and-forth trip over and over again. Successive trips becoming shorter and shorter until the bikes collide and squash the unfortunate bee between the front wheels. What was the total distance travelled by the bee in its many back-and-forth trips from the time the bikes were 20 km apart until its hapless end?

15) A body is moving with a constant acceleration a1 for a distance. It then decelerates with a constant acceleration a2 until it comes to rest. If the total time taken is t and the v-t graph is as shown,

The total distance travelled is

A B

C D

16) Please visit this website for this question: http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html.Using velocity = 100 m s-1, angle = 60o and mass = 10 kg, sketch the path of the motion of the object.

a) Change the mass of the object to 20 kg. Are there any changes in the path of the motion?

b) Introduce air resistance to the motion now by checking the air resistance box. Are there any changes in the path of the motion of the object now?

--------End--------


a1 a2

v

t

http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html

http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html

Kinematics H1/H2

The acceleration of free fall is determined by timing the fall of a steel ball photoelectrically. The ball passes X and Y at times tx and ty after release from P. Show that the

acceleration of free fall is .


h

aP

X

Y

Light beam

Light beam

Point of release

Photo-cells

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