Chapter 2 Kinematics in One Dimension. Mechanics: Study of motion in relation to force and energy,...

Chapter 2

Kinematics in One Dimension

Mechanics: Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object.

Kinematics – Mechanics that describes how an object moves, no reference to the cause (force) or mass of the object. [Gk: kinema = movement].

Dynamics– Mechanics that deals with force: why objects move.

• To describe motion of an object, we need to specify its position, displacement, velocity and acceleration.

• Chap 2: Motion in one dimension only – along a straight line.

• Horizontal direction: position = x, displacement x, velocity vx, acceleration ax.

• Vertical direction: position = y, displacement y, velocity vy, acceleration ay.

• Chap 4 will discuss kinematics in 2-dimension [circular motion, projectile motion].

Introduction

0 x

+

- +

-

y

0 x

+

- +

-

y

• Consider a car driven at a perfectly steady 60 miles per hour on a long and straight section of the highway.

• In the first hour it will travel 60 mi. In the second hour it will cover another 60 mi. etc. We say the car is moving with uniform motion.

Uniform Motion

Uniform Motion

Uniform motion is a straight line motion where:

• Equal displacements are covered in any successive equal intervals of time.

• The velocity is constant.

• The graph of position versus time is a straight line.

Uniform Motion

t

x

x

x

1 2 3 4

60

120

180

240

00

position

velocity

t

v

1 2 3 4

20

40

60

80

00

Non-uniform Motion• An object in non-uniform motion – means

velocity does not stay uniform (constant) during the motion.

• Either its magnitude or direction changes.

• Moving along a straight line with varying speed, or moving along a curved path.

x(t)

t t

v(t)

Average Velocity

• Velocity is the rate of change of position with time.

• The average velocity is displacement divided by the change in time.

vave = x/t

vave = slope of position vs time graph

vave = (xf – xi)/(tf – ti)x

t

x

t

• Instantaneous velocity is the limit of average velocity as t gets small.

• It is the slope of the tangent line to the graph of x(t) versus t.

• It is the derivative of x(t) with respect to t.

dt

dx

t

xlimv

0Δtinst

Instantaneous Velocity

dt

dg

dt

dfg)(f

dt

d

nctdt

df(t)then,ctf(t) If 1nn

Graphs for Instantaneous Velocity

x

tt1 t3

v3>0

v1<0

v2=0

t2

Determine the velocity of the car at times A, B, C and D.

A Positive Zero Negative

B Positive Zero Negative

C Positive Zero Negative

D Positive Zero Negative

x(t)

t

A

B C D

To find position from velocity

f

i

t

t

if

0Δtinst

vdtxx

t

xlim vFrom

dt

dx

xf = xi + Area under velocity-time curve between ti and tf

Final position:

The position of an object changes with time according to the expression

x = t2 + 3t - 9

(a) What is the velocity of the object at t = 3 s?

(b) What is the position of the object when its velocity is 11 m/s?

Example

Area Under velocity-time Graph

Area under the velocity-time graph = magnitude of the displacement over the time interval.

time (s)

velo

city

(m

/s)

uniform (constant) velocity

t1 t2

Area = v(t2-t1) = (x2 – x1) = x

time (s)

velo

city

(m

/s)

Example

A sprinter starts from rest and runs with a constant acceleration for 6 s before reaching his maximum speed. He then runs at the maximum speed the rest of the way. If he ran the 200 m in 25.0 s, at what speed did he cross the finish line?

Average acceleration =t

v a ave

Motion with Constant Acceleration

• Constant acceleration – straight line motion where the speed changes at a constant rate.

• A car moving north. At time t=2sec, its speed is 4 m/s. At t=4 sec, its speed is 8 m/s. At t=6 sec, its speed is 12 m/s. Every 2 sec, its speed changes by 4 m/s. OR, every 1 sec, its speed changes by 2 m/s. It is moving with constant acceleration:

a = 2 m/s/s = 2 m/s2 north.

• Uniformly accelerated motion – when an object moves with constant acceleration.

Uniform (Constant) Acceleration

• An object moving with uniform acceleration – means acceleration stays uniform (constant) throughout the motion.

• Its magnitude and direction stays the same.

• Its velocity changes uniformly.

• Moving along a straight line.

a = slope

t

a(t)v(t)

t

v

t

Graphical Representationve

loci

ty, v

(m

/s)

Time, t (s)

Slope of the line = average acceleration

O

If the slope is zero, the object is moving with zero acceleration (constant velocity)

Motion with Constant Acceleration

t

v

a = slope of the velocity-time graph

T/F? If the acceleration of an object is zero, it must be at rest.

T/F? If an object is at rest, its acceleration must be zero.

a<0

a>0

Slope of the line = average acceleration = v/t

velo

city

, v

(m/s

)

Time, t (s)O

A B

C D

E

1. When is a = 0?

2. When is a < 0?

3. When is a = maximum?

(A) OA

(B) AB

(C) BC

(D) CD

(E) DE

Kinematic Equations for Const. Acceleration

Consider an object moving with constant acceleration a.

As the object moves, its velocity changes.

a aTime = 0

Initial position = xi

Initial velocity = vi

Time = t

Final position = xf

Final velocity = vf

vi vf

a

Kinematic Equations for Const. Acceleration

• aave = ainst

• Initial time = ti , final time = tf Time interval = t = tf - ti

• Initial position = xi , final position = xf

Displacement x = xf - xi

• Initial velocity = vi , final velocity vf

Change in velocity v = vf - vi

Uniformly accelerated motion: a = const.

Kinematic Equations for Constant Acceleration

vf = vi + atxf = xi +vit + ½at2

vf2 = vi

2 + 2axAverage velocity vave = (vi + vf)/2

Uniformly accelerated motion: a = constant.

Time: Initial = ti, final = tf

Positions: Initial = xi, final = xf Velocity: Initial = vi, final = vf

Example

A train traveling at a constant speed of 22 m/s, comes to an incline with a constant slope. While going up the incline the train slows down with a constant acceleration of magnitude 1.4 m/s2.

• Draw a graph of vx versus t.

• What is the speed of the train after 8.0s on the incline?

• How far has the train traveled up the incline after 8.0 s?

Free Fall• Free fall: Only force of gravity acting on an

object making it fall.

• Effect of air resistance is negligible. Motion in vacuum.

• Motion of objects in air where air resistance is negligible (not feather).

• Force of gravity acting on an object near the surface of the earth is F = W = mg.

• Acceleration of any object in free fall:

a = 9.8 m/s2 down (ay = -9.8 m/s2).

• g = 9.8 m/s2 hence ay = -g

Free Fall

+y

+x

ay = -9.8 m/s2

ax = 0

Free Fall contd…

1. ay = -g, regardless of mass of object.

2. ay = -g, regardless of initial velocity

+y

+x

ay = -9.8 m/s2, ax = 0

v0 = 0 v0 = -15 m/s v0 = +15 m/s

3. Free Fall: Motion is symmetric.

+y

+x

ay = -9.8 m/s2, ax = 0

v0 = +5 m/s

At the maximum height:

• vy = 0

• Speed at equal heights will be equal.

• Equal time going up and down.

Example:

A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s and the stone is 1.5 m above the ground when launched.

(a) How high above the ground does the stone rise?

(b) How much time elapses before the

stone hits the ground?

In the figure above showing a graph of position (x) versus time (t), when is the object

(A) at rest?

(B) moving to the left?

(C) moving fastest?

(D) moving slowest?

A car moves at a constant velocity of magnitude 20 m/s. At time t = 0, its position is 50 m from a reference point. What is its position at

(i) t = 2s? (ii) t = 4s (iii) t = 10s?

Example

A stray dog ran 3 km due north and then 4 km due east. What is the magnitude of its average velocity if it took 45 min?

Example

Which position-versus-time graph represents the motion shown in the motion diagram?

Which position-versus-time graph goes with the velocity-versus-time graph at the top? The particle’s position at ti = 0 s is xi = –10 m.

• The average acceleration is the change in velocity divided by the change in time.

• SI unit = m/s2

• Slope of velocity-time graph.

t

v

tt

vv

tt

vv

timeinchange

velocityinchangea

ifif

if

12

t

v(t)

t

vA

C

EB

D

Acceleration (a)

The instantaneous acceleration as at a specific instant of time t is given by the derivative of the velocity

Instantaneous Acceleration

To find velocity from acceleration

f

i

t

t

instif

0Δtinst

dtavv

t

vlima From

dt

dv

vf = vi + Area under acceleration-time curve between ti and tf

OR graphically:

If we know the initial velocity, vi, and the instantaneous acceleration, ainst, as a function of time, t, then the final velocity can be obtained:

Finding velocity from acceleration

Finding velocity from acceleration

Example

The velocity of an object varies as

v(t) = 2t2 - 10t + 18

(a)Find the acceleration of the object at t = 3 s

(b) At what time is the velocity of the object 9 m/s

(c) What is the displacement of the object from t = 0s to t = 2s?

A car moving south slows down with at a constant acceleration of 3.0 m/s2. At t = 0, its velocity is 26 m/s. What is its velocity at t = 3 s?

A. B. C. D. E.

0% 0% 0%0%0%

A. 35 m/s south

B. 17 m/s south

C. 23 m/s south

D. 29 m/s south

E. 17 m/s north

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21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

A car initially traveling at 18.6 m/s begins to slow down with a uniform acceleration of 3.00 m/s2. How long will it take to come to a stop?

A. B. C. D. E.

0% 0% 0%0%0%

A. 55.8 s

B. 15.6 s

C. 6.20 s

D. 221.6 s

E. None of these

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21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise?

A. B. C. D. E.

5% 3%8%

73%

13%

A. 9.8 m

B. 3.00 m

C. 1.15 m

D. 26.1 m

E. 19.6 m

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise?

A. B. C. D. E.

8% 8%10%

69%

5%

A. 9.8 m

B. 3.00 m

C. 1.15 m

D. 26.1 m

E. 19.6 m

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

1. A car initially traveling at a velocity vo begins to slow down with a uniform deceleration of 1.20 m/s2 and comes to a stop in 26.0 seconds. Determine the value of vo.

A. B. C. D. E.

95%

5%0%0%0%

A. 31.2 m/s

B. 21.7 m/s

C. 27.2 m/s

D. 24.8 m/s

E. None of these

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the driver’s reaction time of 0.5 s. What is the minimum stopping distance for the same car traveling at a speed of 40 m/s?

Example