Course Name : Physics – ICourse # PHY 107

Note-1: Introduction, Measurements andOne dimensional motion

Abu Mohammad KhanDepartment of Mathematics and PhysicsNorth South Universityhttps://abukhan.weebly.com

Copyright: It is unlawful to distribute or use this lecture material without the written permission from

the author

Topics to be studied

◮ Physical quantities

◮ Base units and dimensions

◮ Significant figures

◮ One dimensional motion

◮ Geometrical Interpretation of instantaneous and average velocities

◮ Geometrical Interpretation of Acceleration

◮ Equations of Motions

◮ Integral form of Displacement and Velocity

◮ Free Fall: Definition and Properties

◮ Examples

◮ Suggested Problems

Physical Quantities:

Name Symbol Units (SI) Physical Dimensions

Mass m kg M

Length x m L

Time t sec T

1. These are called base quantities because any other physical quantities can beexpressed in terms of these.

2. As for example, the force is expressed symbolically by the letter F and its unit is N(for Newton). This is not the base unit because

Force = Mass × accelerationUnit of F = Units of (Mass × acceleration)

⇒ 1 N = 1 kg.m/s2

Dimension of F = Dimensions of (Mass × acceleration)⇒ [ F ] = [MLT−2]

Significant figures:

◮ This feature is related to the accuracy level of a measurement.

◮ As for example: the measurement of masses 2kg and 2.0 kg are not same. Themeasuring method used for 2.0 kg measurement can measure one-tenth of akilogram accurately, whereas the instrument used to measure 2kg can not.

◮ The fraction of a kilogram has been rounded off, whereas for the other methodthe fraction of one-tenth of a kilogram is rounded off. So the method orinstrument used to measure 2.0 kg is more accurate.

◮ Hence more decimal points implies smaller fraction is measured, and the nextfraction is rounded off. This is known as the significant figure of a measured value.

◮ To find out the significant figure, each number must be expressed in the exponentform:

Any Measured Value or Data Value = (a.bcd · · · )× 10±# .

◮ The number of digits inside the parentheses without the exponential factor is thesignificant figure of the number.

◮ As for example: 1.60 × 10−31 is a three significant number, whereas 1.6 × 10−31

is a two significant number.

◮ Another example: 0.002357 is a four significant number because

0.002357 = 2.357 × 10−3 .

◮ The sig. fig. rules for the arithmetic operations of addition, subtraction, productand division of numbers ate as follows:◮ For addition, subtraction and product, the result will be expressed in the smallest

significant figures, where as for division, the result will be expresses in the largestsignificant figures.

◮ As for example:(a) 2.35 × 3.6792 = 8.64612 ≈ 8.65 (rounded off to 3 sig. fig)(b) 2.35 ± 3.6792 = 6.0292 or -1.3292 ≈ 6.03 or -1.33 (rounded off to 3 sig. fig)(c) 2.35 ÷ 3.6792 = 0.63872581 ≈ 6.3873× 10−1 (rounded off to 5 sig. fig)

◮ This is the scientific method of writing and expressing any numbers. This needsto be followed for the rest of the course and also during your lab classes (if any).

One Dimensional Motion:

◮ One Dimension means that there is only one coordinate, and conventionally it isthe x-axis (horizontally) or y-axis for vertical motion (like free-fall motion, to becovered later on).

◮ The direction of a vector quantity is given by the sign of the parameter whether itis positive or negative. By convention, to the right is positive and to the left isnegative directions.

◮ Similarly, upward is positive and downward is negative.

◮ Definitions of physical quantities:

1. Position: x(t) = Coordinate x as function of time t = (t, x(t)).2. Displacement, ∆x ≡ x(t)− x(t0) = x − x0, where t0 is the initial

time, and t is any later time.

3. Distance, d =

{

Length of the Path|∆x | (Magnitude of the Displacement)

.

4. Instantaneous velocity (or simply velocity): It is defined as,

v(t) ≡dx(t)

dt=

dx

dt= First Derivative of the Position curve

5. Average velocity: It is defined as

vav =∆x

∆t=

x − x0

t − t0(Definition)

= Slope of the straignt line that passes through

the points (t, x) and (t0, x0)

6. Speed: It has two meanings defined by

v =

|vav| = Magnitude of the Average velocity

d

t=

Length of the PathTime

7. The instantaneous acceleration is defined as

a(t) =dv

dt= First derivetive of the velocity graph at time t

= Slope of the tangent line on the velocity graph at time t

=d2x

dt2= Second derivative of the position graph at time t

= Curvature of the position graph at time t

◮ Note that the acceleration is NOT constant in general, and is a function of time.◮ It just gives how fast the velocity itself is changing. If the velocity does not

change (means constant), the a = 0.◮ DO NOT CONFUSE ACCELERATION WITH VELOCITY. THESE ARE

COMPLETELY DIFFERENT !!!◮ Note that, in general, the average velocity between the points b and e is not

equal to the average velocities at points b and e, i.e.

vav =∆x

∆t=

xe − xb

te − tb6=

vb + ve

2.

The position vs. time graph:

vav= Slope of the line

between points b and e= Average velocity

v = Slope of the Tangent linee at point e (dotted line)

td

e

a

b

x(t) vb= Slope of the Tangent line at point b (dotted line)

c

Note the following:

◮ va = vc = ve = 0,because the slope of thedotted line is zero atthese points.

◮ The curvature at point ais positive, because

aa =d2x

dt2> 0 .

◮ But the curvature at d iszero, because the line isstraight through thepoint d .

Some properties of motion:

Based on the mathematical properties of the first and second derivative, it can beconcluded that:

◮ Speeding up implies that either (v > 0, a > 0) or (v < 0, a < 0). That is, bothvelocity and acceleration must have the same sign, either both positive or bothnegative.

◮ Slowing down implies that either (v > 0, a < 0) or (v < 0, a > 0). That is,velocity and acceleration must have opposite signs.

◮ Rest (actual rest) means that both v = 0 AND a = 0.

◮ Momentarily at rest (also known as the ‘turning point’): This requires that ONLYv = 0, BUT a 6= 0.

d

a

b

c

t

x(t)

Note the following:

◮ Note that the points a and c are turning point,because velocities are zero, but acceleration arenot zero.

◮ The velocities changes directions at the turningpoints, from positive to negative or negative topositive directions.

◮ From a → b, it is speeding up, because velocityand acceleration are positive.

◮ From b → c , it is slowing down, because thevelocity is positive, but the acceleration isnegative.

◮ Again from c → d , it is speeding up, becauseboth the velocity and the acceleration arenegative.

Equations of Motion: Constant acceleration:

◮ Since a=constant, theposition x must be aquadratic function of time t,or linear function of time.

◮ The position x can not be afunction of time with negativepowers.

◮ The position can also not bea function of time withfractional powers.

◮ Otherwise the acceleration,being a double derivative ofposition with respect to time,will not be a constant.

Note the following:

◮ Let’s choose the initial time equals zero, i.e.t0 = 0.

◮ From the definition of average velocity, we find:x = x0 + vavt = x0 + vt .

◮ From the definition of average acceleration, wealso find,v = v0 + aavt = v0 + at = v0 + at . Note thatfor constant acceleration, aav a = a.

◮ Using v = (v + v0)/2 which is the arithmeticmean of velocities, we also get,x = x0 + v0t +

12at2 .

◮ Finally, removing time t from the previousexpression, we also get,v2 = v20 + 2a(x − x0) .

Short Summary:

◮ Only for constant acceleration, the equations of motions are:

(1) x = x0 + vt.(2) v = v0 + at.

(3) x = x0 + v0t +1

2at2.

(4) v2 = v20 + 2a(x − x0).

◮ Only for constant acceleration, the average velocity can also be expressed as thearithmetic average of two velocities at the final and initial times, i.e.

v = vav =

∆x

∆t(Definition. So always valid)

v + v0

2((Valid iff acceleration is constant)

Integral Form of Displacement and Velocity:◮ From the definition of velocity, we write:

v =dx

dt⇒ dx = vdt Integrating :

∫ x

x0

dx =

∫ t

t0

vdt ,

∆x ≡ x − x0 =

∫ t

t0

vdt = Area (A) under the curve in v -t graph

= v∆t (iff v is constant)

◮ Note in the v-t graph that: A1 > 0, A2 > 0, A4 > 0 and A3 < 0. Therefore, thedisplacements are:

xb − xa = A1 ,

xc − xb = A2 + A4 − |A3| ,

and xc − xa = (A1 + A2 + A4)− |A3| .t

v(t)

a cb

Areas are positive above the t−axis

Area is negative below the t−axis

A1A2

A3

A4

The Acceleration vs. Time Graph:

◮ Similarly, from the a-t graph, it is easy to see that the areas are: A1 > 0, A2 > 0and A3 < 0.Therefore, the changes in velocity are:

vb − va = A1 ,

vc − va = A1 + A2 ,

and vd − va = (A1 + A2)− |A3| .

A1 A2

A3

d

a(t)

tc

b

a

◮ Note that when the acceleration is constant, the graph is horizontal. So, theregion under the curve is perfectly rectangular, and the area under the curve is(a)(∆t), where a is the height of the rectangle and ∆t is the width of therectangle.

Free Fall:

◮ The free fall is an example of one-dimensional motion.

◮ The position is represented by the vertical axis (the y -axis). So, the notation hasbeen changed accordingly.

◮ The conditions to be satisfied for any Free-Fall motion are:

1. The acceleration is given by: a = −g = −9.80m/s2. Therefore,g = |a| = 9.80m/s2.

2. The acceleration is fixed (both in magnitude and direction) by the Gravitationalforce, NOT by any other means.

◮ Therefore, the equations of motion for Free-Fall become:

(1) y = y0 + vt.(2) v = v0 − gt.

(3) y = y0 + v0t −1

2gt2.

(4) v2 = v20 − 2g(y − y0).

Free-Fall Properties:

◮ The time to go up equals the time to go down if thedisplacement is zero:

tup = tdown =v0

g(v0 = initial speed)

◮ At any level ( i.e. same height from the ground):

vup = −vdown =⇒ |vup| = |vdown| .

◮ The total time of the flight (when displacement equalszero) is

T = 2tup = 2tdown =2v0g

.

◮ At the max. height, the velocity changes direction, butthe acceleration is still non-zero constant. Hence, it is aturning point.

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tup tdown

vup

Max. Height(Turning point)

y0 = y (Ground)

vdown

v=0, a = −g

v0 v= −v0

Example:

The position of an object moving along an x axis is given by

x = 3t − 4t2 + t3 ,

where x is in meters and t in seconds. Answer the following:

(a) The object’s displacement between t = 0 to t = 4s.;

(b) What is its average velocity for the time interval from t = 2s to t = 4s?

(c) Find the turning point of the object.

(d) When will the acceleration be zero?

Solution:(a) The displacement is ∆x = x4 − x0 =

[

(

3× 4− 4× 42 +43)

− (0)]

m = 12m . X

(b) The average velocity is

vav =∆x

∆t=

x4 − x2

t4 − t2=

(3× 4− 4× 42 + 43)− (3× 2− 4× 22 + 23)

4− 2m/s = 7m/s .X

(c) At the turning point: v = 0. Therefore, v = dx/dt = 3t2 − 8t + 3 = 0. Thesolution is

t =−(−8)±

√

(−8)2 − 4× 3× 3

2× 3sec = 0.45 sec or2.2 sec .X

These are when the turning points occur. The location (or position) where theturning points occur are:

x(t = 0.45) =[

3(0.45) − 4(0.45)2 + (0.45)3)]

m = 0.63m ,X

and x(t = 2.2) =[

3(2.2) − 4(2.2)2 + (2.2)3)]

m = 2.1m .X

(d) The acceleration is: a =dx2

dt2= −8 + 6t. Therefore, a = 0 yields t = 1.33 sec X

Problem # 2.34

In the figure below, a red car and a green car, identical except for the color, movetoward each other in adjacent lanes and parallel to an x axis. At time t = 0, the redcar is at xr = 0 and the green car is at xg = 220m.

If the red car has a constant velocity of 20 km/h, the cars pass each other atx = 44.5m, and if it has a constant velocity of 40 km/h, they pass each other atx = 76.6m. What are:

(a) the initial velocity and

(b) the constant acceleration of the green car?

Solution # 2.34

The schematic diagram is shown in theadjacent figure. Here, v0 and a areunknown.

◮ In the Case-I, we need to find t

for the red car, and write anequation for the green car(because t is same for both cars).

◮ In the Case-II, again we find t forthe red car, and write an equationfor the green car (same reason asbefore, but for different distance).

◮ Thus, we will get two equationswith two unknown and solve byusing the method of elimination.

950

t=0

v0

att

x = 44.5 m(Cars meet here after time t)

Case−I:

v=20km/h = m/s

(Cars meet here after time t)

t=0

v0

a

Case−II:

9100

t t

x = 76.6 m

v=40km/h = m/s

◮ Case-I: Suppose after time t the cars meet at a position ( or distance) 44.5m.

Since, v = 20 km/h = 50/9m/sec, the time t is t =x

v=

44.5 × 9

50sec = 8.0 sec .

Now, in time t = 8.0 sec, the green car’s position satisfy (3rd equation of themotion)

44.5 = 220 + 8v0 +1

2a(8)2 =⇒ 32.0a + 8.0v0 + 175.5 =0 .

◮ Case-II: In this case, the cars meet at a position (or distance) 76.6m. Since,

v = 40 km/h = 100/9m/sec, the time t is t =x

v=

76.6 × 9

100sec = 6.9 sec . Now,

in time t = 6.9 sec, the green car’s position satisfy (3rd equation of the motion)

76.6 = 220 + 6.9v0 +1

2a(6.9)2 =⇒ 23.8a + 6.9v0 + 1143.4 = 0 .

◮ Solving these equations (red colored equations) for initial velocity and accelerationusing the method of elimination, we find that

Initial velocity of the green car, v0 = −13.5m/s .X

The acceleration of the green car, a = −2.1m/s2 .X

Problem:

A baseball is tossed up with initial speed of 12m/s.

(a) How long does it take to reach the maximum height?

(b) Find the maximum height.

(c) How long dies it take to reach 5.0m above the releasepoint?

(d) What is the velocity at 5.0m above the release point?

Solution:

Here the ground is the release point: y0 = 0.

(a) Here v = v0 + at gives:0 = 12− 9.8tup ⇒ tup = (12/9.8)sec = 1.22 sec. X

(b) In this case, we have, v2 = v20 + 2a(y − y0) = 0. Therefore,we find,

h = |y − y0| =v202g

=(12)2

2× 9.8m = 7.35m .X

y0=0

v0 = 12 m/s

5.0 m

Max height (v=0)

v

h

tup

(c) Here y = 5.0m. Therefore, the 3rd equation of motiongives,

y = y0 + v0t +1

2at2 ,

5 = 0 + 12t −1

2(9.8)t2 ,

0 = 4.9t2 − 12t + 5 ,

t =−(−12)±

√

(−12)2 − 4× 4.9 × 5

2× 9.8sec ,

= 0.53 sec , 1.91 sec . X

Note that 0.53 sec is the time when the velocity is positive(that is it, going upward), and 1.92 sec is the time when thevelocity is negative (that is, going downward).

y0=0

v0 = 12 m/s

5.0 m

Max height (v=0)

v

h

tup

(d) In the previous part, we found two times. That is thebaseball was at the height of 5m from the ground twice: at0.53 sec and 1.92 sec. The velocities at these two times are:

v(t = 0.53 sec) =(

v0 + at)

t=0.53 sec,

=(

12− 9.8× 0.53)

m/s ,

= 6.81m/s . X

v(t = 1.92 sec) =(

v0 + at)

t=1.92 sec,

=(

12− 9.8× 1.92)

m/s ,

= −6.81m/s . X

y0=0

v0 = 12 m/s

5.0 m

Max height (v=0)

v

h

tup

Suggested Problems:

• Chapter-1: 3, 9, 12, 20, 27.• Chapter-2: 3, 5, 7, 15, 18, 19, 21, 25, 28, 34, 36, 41, 44, 48, 54, 55, 65, 67, 69, 70.