Kinematics in One Dimension

PHY130 – FUNDAMENTAL PHYSICS I CHAPTER 2 KINEMATICS IN ONE DIRECTION

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Lesson Contents

2.1 Scalars and vectors 2.2 Linear motion parameters 2.2.1 Definition of linear motion parameters 2.2.2 Average and instantaneous velocity 2.2.3 Average and instantaneous acceleration

2.3 Graph of linear motion 2.3.1 Displacement – time graph

2.3.2 Velocity – time graph 2.4 Linear motion with constant acceleration

2.5 Free fall motion

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Lesson Outcomes

After studying this chapter, you should : • Able to define & write equation for the motion parameter: displacement, velocity and acceleration

• Able to explain & differentiate between the average & instantaneous values of the velocity and acceleration • Able to determine the motion parameters from the displacement-time graph and the velocity-time graph • Able to explain and analyze motion with constant acceleration • Able to explain, analyze and solve falling object problems 3

2.1 SCALARS AND VECTORS

Scalar & Vectors

Each physical quantity can be categorized as either a scalar quantity or a vector quantity.

QUANTITIES

Scalar

Vector

DEFINITION

physical quantities that have magnitude only

physical quantities that possess both magnitude and direction.

EXAMPLES

• Distance • Work • Charge • Speed • Power

• Displaceme • Momentum nt • Velocity • Electric field • Force

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2.2 LINEAR MOTION PARAMETERS

Distance & Displacement

DISTANCE

DISPLACEMENT

DEFINITION

total path length the shortest distance from the travelled from one initial position to the final position location to another

QUANTITY

scalar

vector

SI UNIT

metre (m)

metre (m)

VALUE

Always positive

1. Either Positive, negative or zero • Positive : right ( ) • Negative : left / opposite ( ) 2. In calculating displacement = Final position – Initial position

x = xf - xi 7

Speed & Velocity

SPEED

VELOCITY

DEFINITION

How fast an object travels in a given time interval

rate of change of displacement

QUANTITY

Scalar

Vector

SI UNIT

m/s @ ms-1

m/s @ ms-1

FORMULA

Distance / Time

Displacement / Time interval **direction of velocity = direction of change in displacement

TERMS

Instantaneous Speed : Speed at any instant

Instantaneous velocity : Velocity at any instant

Constant / uniform speed : Equal distance moved in equal time intervals

Constant / uniform velocity : Velocity does not increase or slow down

Average speed 

total distance moved total time taken

Average velocity 

total displacement total time taken 8

Acceleration

ACCELERATION DEFINITION

A change of velocity over time interval

QUANTITY

Vector

UNIT

m/s2 @ ms-2

TERMS

Deceleration = Negative acceleration

Instantaneous acceleration : Acceleration at any instant. constant acceleration : velocity increases @ decreases by the same amount for every second of its motion.

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s

Relationship of velocity and displacement

Gradient = constant speed = gradient

0

(a) Uniform velocity

t

s

s Q

(c)

P

R

Gradient at point R is negative.

Gradient at point Q is zero.

0

t

Gradient increases with time

The direction of velocity is changing.

The velocity is zero.

0 (b) The velocity increasest with time

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Relationship of velocity and acceleration

Area under v-t graph =

displacement/distance Gradient = acceleration

v

v

v

B

Uniform acceleration

Uniform velocity

C

A 0

t1 (a) t2

t

0

t1

(b) t2

t

0

t1

t2 (c)

t

2.3 GRAPHS OF LINEAR MOTION

EXAMPLE A toy train moves slowly along a straight track according to the displacement, s against time, t graph in figure below.

s (cm)

(i) Displacement - time graph

10

8 6 4 2 0

2

4

6

8

10 12 14

t (s)

TIPS!!You must know to: 1. Sketch a graph from the given information or vice versa 2. Calculate the gradient (velocity/speed), average velocity and instantaneous velocity

a. Sketch a velocity (cm s-1) against time (s) graph. b. Determine the average velocity for the whole journey.

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

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SOLUTION a)

v (cm s1)

s2  s1 b) vav  t 2  t1

1.50

0.68 0

2

4

6

8

10 12 14

t (s)

10  0 vav  14  0 v a v  0 .7 1 4 c m s 1

c)v  average velocity from 10 s to 14 s s2  s1 10  4 v  t2  t1 14  10

 1 .5 0 c m s

1

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

v (m s 1) (ii) Velocity - time graph

4 2 0 -2 -4

5

10 15

20 25 30 35 40 45

50

t (s)

TIPS!!You must know to: 1. Sketch a graph from the given information or vice versa 2. Calculate the gradient (acceleration) or area of graph (displacement/distance)

a. Sketch a graph of acceleration (m s-2) against time (s). b. Determine the total distance traveled by the lift and its displacement. c. Calculate the average acceleration between 20 s to 40 s.

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SOLUTION a (m s2)

a)

a.

0.8 0.6 0.4 0.2 0 -0.2

5

10 15

20 25 30 35 40 45

50

t (s)

-0.4

-0.6 -0.8 16

SOLUTION b)i

TIPS!!

1.Total distance = total area of the graph (both + Y axis and –Y axis)

v (m s 1) 4 2 0 -2

A1 5

A2 10 15

A3

2. Total displacement • area at + Y axis = +ve area • Area at –Y axis = -ve area

20 25 30 A 35 40 45 4 A5

50

t (s)

-4

Total distance  area under the graph of v-t  A1  A 2  A3  A 4  A5 1 1 1 1 1 Total distance  25  2  410  5  104  54  15  54 2 2 2 2 2

T o ta l d is ta n c e  1 1 5 m

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SOLUTION Displacement  area under the graph of v-t

b) ii

 A1  A 2  A3  A 4  A5 

c)

1 25  1 2  410  1 5  104  1 5 4  1 15  5 4 2 2 2 2 2

D is p la c e m e n t  1 5 m v2  v1 aav  t 2  t1 44 aav  40  20 a a v   0 .4 m s 2

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2.4 LINEAR MOTION WITH CONSTANT ACCELERATION

REMEMBER!!

The 4 equations

TIPS!!

Where

v = u + at s = ut + 2

1 2

at

2

2

v = u + 2as s=

1 2

(u + v)t

v = final velocity u = initial velocity a = acceleration t = time s = displacement

JUES

Word  Symbol  numbers 1. Read the question carefully.

2. JOT DOWN the given information using symbols and value.

Eg : acc, a = 10 ms-2 3. What is the unknown value? HINT WORD : calculate, determine.

4. Write in symbols.

Eg : final velocity, v= ??

5. Select the most suitable equation 6. Solve it. GOOD LUCK!!

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EXAMPLE 1. A box starts from rest and moves with a speed of 2.2 ms-1 in 3 seconds.

Calculate the : [MARCH 2014] (i)

acceleration of the box

(ii) distance it moved in the first 6 seconds.

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SOLUTION i)

Acceleration of the box v = u + at

s = ut +

2.2 = 0 + a(3) a = 0.733 ms

ii) Distance it

s =0+ -2

1 2 1

2

moved in the first 6 seconds.

at

2

(0.733)(6)

s = 13.194 m

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2.5 FREE FALL MOTION

1) 2)

The direction of motion are now along the vertical y-axis.

Vector quantities (displacement, initial velocity & final velocity) are positive in an upward motion & negative in a downward motion. + s y, + u y, + v y reference level

- s y, - u y, - v y

3)

The free-fall acceleration is always negative in an upward & a downward motion.

- g = - 9.81 m s

-2

- g = - 9.81 m s

-2

reference level

4)

t is always positive.

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TRANSITION

PREVIOUS

The 4 equations

Where

v = final velocity u = initial velocity a = acceleration g = gravitational acceleration t = time s = displacement

v = u + at s = ut + 2

1 2

at

2

2

v = u + 2as s=

1 2

NOW

TIPS!!

1) a ďƒ -g = - 9.81 ms-2

v = u - gt s = ut -

2) Max height : v = 0 ms-1 3) Reference point : s = 0 m

2

+ s y, + u y, + v y, + t reference level

2

gt

2

2

v = u - 2gs

4)

(u + v)t

1

s=

1 2

(u + v)t

- s y, - u y, - v y, + t 25

EXAMPLE A book is dropped 150 m from the ground. Determine a. the time taken for the book reaches the ground. b. the velocity of the book when it reaches the ground. (given g = 9.81 m s-2)

uy = 0 m s1 Hence 150 m

1 2 s y  u y t  gt 2 1  150  0  9.81t 2 2

t  5.53 s

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SOLUTION a. The vertical displacement is sy = 150 m

1 2 s y  u y t  gt 2 1  150  0  9.81t 2 2 t  5.53 s

SOLUTION b. The book’s velocity is given by

uy  0

vy  u y  gt

v y  0  9.81 5.53  1 v y   54.2 m s OR

s y   150 m

v y  u y  2 gs y 2

2

v y  0  2 9.81 150  2

v y  ? v  54.2 m s 1 ( downward ) y Therefore the book’s velocity is

v y   54.2 m s

1 28