chapter-1

Chapter 1Motion in a straight line

1-2 Displacement vs Distance Average Velocity1-3 INSTANTANEOUS VELOCITY

1 - 4 Acceleration1.6 the acceleration of gravity and falling objects

10/19/2011 1Norah Ali Almoneef

1-2 Displacement vs DistanceAverage Velocity

• Displacement is a vector that points from an object’s initial position to its final position

• and has a magnitude that equals the shortest distance between the two positions.

_Only depends on the initial and final positions– Independent of actual paths between the initial and

final positions• Distance is a scalar

– Depends on the initial and final positions as well as the actual path between them


3

Displacement

x (m)

t (s)4

3

-3

Displacement between t=1 s and t=5 s

x = 1.0 m - 2.0 m = -1.0 m

This type of x(t) plot shows the position of an object at any time, e.g.,

Position at t=3 s, x(3) = 1 m

10/19/2011 Norah Ali Almoneef

Given the train’s initial position and its final position what is the displacement of the train?

What is the distance traveled by the train ?Displacement =

f ix x x


Example:A boy travels from D to A,A to B .B to C.C to D

Displacement from D to D ( which are initial and final points ) = 0

Distance traveled = 8 +4+8+4 = 24 m


Example :

Distance = 4 m + 3 m =7 m

Displacement = 5 m


Speed and VelocityThe average speed being the distance traveled divided by the time required to cover the distance:

How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m /s?

Distance = 5400 s x 2.22 m / s = 11988 m


Speed

Speed can be defined in a couple of ways:How fast something is movingThe distance covered in a certain amount of timeThe rate of change of the position of an object

Units for speed are: miles / hour (mi/hr)kilometers / hour (km/hr)feet / second (ft/s)

This is the standard unit meters / second (m/s)




A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m.

Determine the velocity of the particle when t = 4 s.

tt 63 2

At t = 4 s,

the velocity = 3 (4)2 – 6(4) = 24 m/s

dt

dxv

example


What is

example


From A to B

What is

A B

example





1-3 INSTANTANEOUS VELOCITY


Instantaneous velocity – is how fast an object is moving at a particular instant.

The position of a particle moving on the x axis is given by x = 7.8 + 9.2 t – 2.1t2. What is its instantaneous velocity at t = 3.5 seconds?

v = 0 + 9.2 – (2)(2.1)tv = 0 + 9.2 – (2)(2.1)(3.5) = -5.5 m/s

dt

dx

t

xv

t

0

limexample

1 - 4 AccelerationAcceleration: is a rate at which a velocity is changing.

Instantaneous acceleration

= dv / dt = d2 x / d t2


Example

A car’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocity of 26 m/s. What is the acceleration of the car?

The car is increasing its velocity by 8 m/s for every second it is moving.



• Instantaneous Acceleration

• Suppose a particle is moving in a straight line so that its position is given by the relationship x = (2.10 m/s2)t3 + 2.8 m. Find its instantaneous acceleration at 5 seconds.v = dx / dt = (3)(2.1)t2

a = dv / dt = (2)(3)(2.1)t at t= 5sa = (2)(3)(2.1)(5) = 63 m/s2

dt

dv

t

va

t

0

lim

• Instantaneous Acceleration


example


A bullet train starts from rest from a station and travels along a straight horizontal track towards another station. The graph in fig. shows how the speed of the train varies withtime over the whole journey. Determine: (a) the total distance covered by the train,(b) the average speed of the train.

Speed / ms-1

Time

40

02 12 16

example

Average speed = (total distance) / (total time ) = 520 / 16 = 32.5 ms-1

A ) Total distance travelled

mx

x

5208040040

42

04010402

2

040

OR Total distance travelled= ‘area under the graph’= (1/2)(10 + 16)(40) = 520 m


: a car is traveling 30 m/s and approaches 10 m from an intersection when the driver sees a pedestrian and slams on his brakes and decelerates at a rate of 50 m/s2.(a) How long does it take the car to come to a stop?(b) how far does the car travel before coming to a stop?

vf -vi= a t, where vo= 30 m/s, v = 0 m/s, and a = -50 m/s2

t = (0 -30)/(-50) = 0.6 s

Δx= vit + ½ a t2= (30)(0.6) + ½(-50)(0.6)2= 18 -9 = 9 m

example


Equations of Kinematics for Constant Acceleration

1.5 finding the motion of an object


• How long does it take a car going 30 m/sec to stop of it decelerates at 7 m/sec2?

example


- A car starting from rest attains a speed of 28 m/sec in 20 sec. Find the acceleration of the car and the distance it travels in this time.

example



1. Velocity & acceleration are both vectors.Are the velocity and the acceleration always in the same

direction? NO!!

If the object is slowing down, the acceleration vector is in the opposite direction of the velocity vector!

2. Velocity & acceleration are vectors.Is it possible for an object to have a zero acceleration and a

non-zero velocity? YES!!

If the object is moving at a constant velocity, the acceleration vector is zero!



Examples :1 ) What is the acceleration of a car that increased its speed from 10 m/s to 30 m/s in 4 seconds? a = (30 m/s – 10 m/s) ÷ 4s

= 20 m/s ÷ 4s= 5 m/s2

2)the same car now slows down back to 10 m/s in 5 seconds. What is his acceleration? a = (10 m/s – 30 m/s) ÷ 5s

= (- 20 m/s) ÷ 5s= - 4 m/s2

Means slowing down


*deduce from the shape of a speed-time graph when a body is:(i) at rest(ii) moving with uniform speed(iii) moving with uniform acceleration(iv) moving with non-uniform acceleration

Graphical Analysis

Velocity

(i)

Time

(ii)

(iii) (iv)

(i) at rest(ii) moving with uniform speed(iii) moving with uniform acceleration(iv) moving with non-uniform acceleration


example


Velocity /m/s

(i)

(ii)

(iii)

Time/s0 10 14 23

A bus stopped at a bus-stop for 10 seconds before accelerating to a velocity of 15 m/s in 4 seconds and then at a constant speed for the next 9 seconds. How does the graph look like? How far did the bus go in this 23 seconds?

15

• Distance travelled in first 10 seconds is zero• Distance travelled in the next 4 seconds is

= ½ x 4 x 15 = 30 m• Distance travelled in the final 9 seconds is

= 9 x 15 = 135 m• Total distance travelled = 165 m



Objects thrown straight up

The acceleration of gravity (g) for objects in free fall at the earth's surface is 9.8 m/s2. ( down ward )

The acceleration of a falling object is due to the force of gravity between the object and the earth.

Galileo showed that falling objects accelerate equally, neglecting air resistance.

Galileo found that all things fall at the same rate.

On the surface of the earth, in a vacuum, all objects accelerate towards the surface of the earth at 9.8 m/s2.

g actually changes as we move to higher altitudes

1.6 the acceleration of gravity and falling objects


v v0 gt

y v0t 1

2gt 2

v 2 v02 2gy

v v0 at

x v0t 1

2at 2

v 2 v02 2ax

Equations of Kinematics for Constant Acceleration

For free fall



A ball is dropped from a tall building and strikes the ground 4 seconds later. A ) what velocity does it strike the ground B ) what distance does it fall?


g = -


How high can a human throw a ball if he can throw it with initial velocity 90 m / h?.

example


example


example


Notice in free fallNotice in free fall


• “free-fall” acceleration due to gravity a=9.8m/s2, down

• “at rest” not movingv=0

• “dropped” starts at rest and free-fallvi=0 and a=9.81m/s2, down

• “constant velocity” no accelerationa=0

• “stops” final velocity is zerovf=0

Word clues to numbers for problem solving


1.Displacement:

2. Average velocity:

f ix x x

xv

t

3.Instantaneou

s velocity:0t

x dxv Lim

t dt

2. Time interval:

4. Average acceleration:

f iv va

t

5.nstantaneous

acceleration: 0t

v dva Lim

t dt

Summary


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