2.1 Linear Motion

Physics Form 4 Chapter 2 Forces and Motion

Name: ____________________________________ Form: _________ Date: ___________

2.1 LINEAR MOTION

2.1.1 Linear Motion

Linear motion is motion along a straight line.

2.1.1a Kinematics and Dynamics

Kinematics is the study of motion without taking into consideration the forces involved. Dynamics is the study of motion taking into consideration the forces involved.

2.1.1b Important Physical Quantities in Kinematics

Distance, displacement, speed, velocity and acceleration are physical quantities that are important in the study of the motion of an object.

No.Physical Quantity

and SymbolUnit Equation and definition

1Distance

(Scalar)s m

Distance = Speed x Times = vt

Length of the path an object moved.

2Displacement

(Vector)s m

Displacement = Velocity x Time

s = vtDistance moved in a particular direction or distance of an object from a given point in a particular direction.

3Speed

(Scalar)v ms-1

4Velocity

(Vector)v ms-1

5

Acceleration

(Vector)

a ms-2

6Time

(Scalar)t s (Time is involved in all the equations above)

2.1.1c Distance and Displacement


Name: ____________________________________ Form: _________ Date: ___________

Distance and displacement are two different physical quantities. Distance is a scalar quantity that has magnitude only. Displacement is a vector quantity that has both magnitude and direction.

Distance s is the length of the path an object moved.

Distance = Speed x Time OR s = vt

Displacement s is the distance moved in a particular direction or distance of an object from a given point in a particular direction.

Displacement = Velocity x Time OR s = vt

Example 1A student walks 3 km towards the North from point O. After that, he moves 4 towards the East. Calculate (a) the total distance traveled and (b) the displacement of the student from point O?

Solution a. Total distance moved = Distance towards the North + Distance towards

the East = 3 + 4 = 7 km.b. The triangle OAB is a right angle triangle.

To calculate the displacement from O, use Pythagoras theorem:

Displacement = 5 km in the direction of OB. Notice that distance is a scalar quantity; so, total distance

moved is calculated by adding the distances moved in different directions like ordinary numbers.

Displacement is vector. The magnitude of the displacement cannot be calculated by adding the displacements in different directions like ordinary numbers.

Refer to the box to see how the direction of the displacement is determined.

Example 2A student walks 6 m to the right of a point O. He then walks 5 m towards the right. Finally he walks 11 m towards O. Calculate (a) the distance moved by the student and (b) the displacement of the student from O. (a) The distance moved by the student = 6 + 5 + 11 = 22 m

(b) The displacement of the student from O = 0 m (The student has moved back to O!)

2.1.1d Speed and Velocity

4 km

3 km

O

A B

5 km

Determination of the direction of the displacement

= 53o

Direction of the displacement is 053o (stated in bearings).

6 mO 5 m

11 m


Name: ____________________________________ Form: _________ Date: ___________

Like distance and displacement, speed and velocity are different. Speed is a scalar quantity. Velocity is a vector quantity.

Speed v is the rate of change of distance

OR

Velocity v is the rate of change of displacement

OR

Example 2 A car moves 400 km Northwards from the point O in 5 hours and continues to move 300 km Eastwards in 5 hours. Calculate (a) the speed and (b) the velocity of the car? Solution a. Total Distance moved, s = 400 + 300 km.

b. Final displacement of the car from O is 500 km. (Using Pythagoras theorem.)

Example 3 An object moves round a circle with a perimeter of 200 m in 4 s. Calculate (a) the speed of the object and (b) the velocity of the object. Solution(a) Speed of the object, v = 200/4 = 50 m s-1

(b) Velocity of the object, v = 0/4 = 0 m s-1

2.1.1e Uniform velocity and Non-uniform velocity

An object is moving with a uniform velocity if the speed and the direction of motion do not change. This tells us that an object has uniform velocity if it is

300 km

400 km

O

A B

500 km

How is the acceleration of an object related to its speed and direction?

Can you determine the direction of the velocity?

O

When an object moves round a circle it comes

back to its starting point. So its displacement = 0


Name: ____________________________________ Form: _________ Date: ___________

(a) moving in a straight line (no change in direction) with (b) constant speed (no change in speed).

An object is moving with a non-uniform velocity if the speed changes but the direction of motion remains constant. the speed remains constant but the direction of motion changes. both the speed and the direction of motion changes.

2.1.1f Acceleration

An object is accelerating if its velocity is NOT uniform. An object is NOT accelerating if its velocity is uniform.

The Table below summarizes the velocity, speed and direction of motion of an object that is accelerating and not accelerating.

Acceleration Speed Direction of Motion Velocity

Object is accelerating Changes Fixed Changes

Object is accelerating Fixed Changes Changes

Object is accelerating Changes Changes ChangesObject is not accelerating

(Zero acceleration)Fixed Fixed Fixed

Acceleration is the rate of change of velocity

OR

Retardation or DecelerationAn object is undergoing retardation or is decelerating if the speed (magnitude of its velocity) is decreasing.

Uniform AccelerationAn object has uniform acceleration if its velocity increases at a constant rate.

Zero Acceleration An object is said to have zero acceleration if the object is a. at rest or b. moving with uniform velocity (moving with uniform speed in a fixed direction)

Positive Acceleration or Negative Acceleration A positive acceleration does not always show an object is accelerating. Neither does a negative acceleration always show an object is decelerating. o For example, if an object is moving in the negative direction and the acceleration is negative, the

object is accelerating not decelerating.

2.1.2 Linear motion with constant acceleration (‘suvat’ Equations)

‘suvat’ Equations can be used if an object moves in a straight line with constant acceleration. There are 4 equations for linear motion with constant acceleration:


Name: ____________________________________ Form: _________ Date: ___________

Equation Quantity

REMEMBER suvat

Note: There is a fifth equation embedded in the equation: Displacement = Average velocity

x Time s = ½ (u + v)t

v = u + ats = ½ (u + v)t2as = v2 – u2

s = ut + ½ at2

s = Displacementu = Initial velocityv = Final velocitya = Accelerationt = Time

Example 1A car accelerates from 20 m s-1 to 40 m s-1 in 5 s. Calculate the acceleration of the car. Solution

Example 2An object moving with a velocity of 10 m s-1 undergoes a retardation of 5 m s-2 until the object stops. Calculate the distance moved by the object before it stops. Solution

Example 3A bullet moving at 400 m s-1 hits a wooden block which is at rest. The bullet penetrates the block a depth of 4 cm before it stops. What is the deceleration of the bullet?

2.1.3 Analysing motion with a ticker timer

A ticker timer can only work with alternating current (a.c.). The potential difference used is 12 V a.u. and the frequency of the alternating

current is 50 Hz.

Datau = ____ m s-1 v = ____ m s-1

t = ____ s

Datau = ____ m s-1

v = ____ m s-1 (stops)a = ____ m s-2

Why is the acceleration negative?

Data

u = _______ m s-1

v = _______ m s-1

s = ________ cm


Name: ____________________________________ Form: _________ Date: ___________

o The striker of the ticker timer vibrates 50 times a second and produces 50 dots a second on a ticker tape.

2.1.3a Measurement of time with a ticker timer

Time is measured by a ticker timer in units of ticks and NOT dots. Each mark on the ticker tape is a dot. The space between two dots is the time interval between 2 dots, that is, a tick.

o Note: To obtain the time shown on a ticker tape, we have to count the number of spaces in a ticker tape.

The frequency of a ticker timer is 50 Hz. This means that 50 ticks (50 spaces) are produced per second on a ticker tape, that is, o 50 ticks = 1 second

o

The strip of ticker tape below has 11 dots. This means that there are 10 spaces between the 11 dots. So, the time shown by the ticker strip is 10 ticks. o Time for 10 ticks in seconds = 10 x 0.02 = 0.2 s

2.1.3b Measurement of Displacement or Distance with a Ticker Timer

Displacement or Distance of an object = Length of the tape between the starting point A and the final point B.

NOTE: The ticker tape moving through a ticker timer can only move in straight line and in one direction. So, the distance measured by a ticker is the displacement because it is distance moved in a fixed direction.

2.1.3c Measurement of Velocity with a Ticker Timer

Dots

Tick

First dot

Eleventh dot

There are 10 spaces between 11 dots! Note: Time in ticks is measured in terms of the number of spaces

and NOT the number of dots.

A B

Length of tape

s = 20 cm

t = 4 ticks = 0.08 s

A B


Name: ____________________________________ Form: _________ Date: ___________

The displacement between point A and point B is 20 cm and the time for the object to move from A to B is 0.08 s. Velocity can be using the equation:

2.1.3d Qualitative Analysis of Velocity with a Ticker Timer

The two strips of ticker tapes for the motion of two objects. The distance between consecutive dots for each strip is uniform. This shows that both objects are moving with uniform velocity.

However, the distance between consecutive dots for strip Q is larger compared to that of strip P. This tells use that object Q is moving with a higher velocity.

ExampleDescribe the motion of objects A, B, and C based on the strips of ticker tapes below:

Solution Tape A: Object A moves with a (uniform/non-uniform) velocity. Initially, the velocity

of the object is (high/low). The velocity of the object is (increasing/ decreasing) continuously.

Tape B: Object B moves with a (uniform/non-uniform) velocity. Initially, the velocity of the object is (high/low). The velocity of the object is (increasing/ decreasing) continuously.

Tape C: Initially, object C moves with (uniform/non-uniform) velocity. At the fifth tick, its velocity is (half of/the same as) the original velocity. At the eleventh tick, the velocity is back to the initial value and is (uniform/non-uniform).

2.1.3e Measurement of Acceleration using a Ticker Timer

A velocity-time chart is obtained by arranging strips of ticker tape of equal time interval (measured in ticks). Strips with 10 ticks are usually used which means that the time for each strip is 0.2s. Since the time for each strip is the same, a comparison of the length of the strip will give a comparison of the velocity.

Start EndA

B

C

P

Q


Name: ____________________________________ Form: _________ Date: ___________

2.1.3f Qualitative analysis of a velocity-time chart

The time interval for a strip in each of the velocity-time charts A – F is 10 ticks. a. What is represented by the x and y-axes?

x-axis: y-axis:

b. Comment on the velocity and the acceleration of the motion of the object represented by each chart?

Solution Chart A: Each strip is of (the same/different) length. The velocity of the object is (uniform/non-

uniform). The acceleration is (zero/constant/not constant). Chart B: Each strip is of (the same/different) length. The velocity of the object is (uniform/non-

uniform). The acceleration is (zero/constant/not constant). The strips in Chart B are (shorter/longer) than those in Chart A showing that the velocity of object B is (lower/higher).

Chart C: Initially, the velocity of the object is (low/high). The velocity of the object (decreases/increases) uniformly. The object has (uniform/non-uniform) acceleration.

Chart D: Initially, the velocity of the object is (low/high). The velocity of the object (decreases/increases) uniformly. The object has (uniform/non-uniform) deceleration.

Chart E: Initially, the velocity of the object is (zero/uniform/non-uniform). Subsequently, the velocity (decreases/increases) uniformly. The object has (zero/uniform/non-uniform) acceleration initially. After that, it undergoes uniform (retardation/acceleration).

Chart F: Initially, the velocity of the object (decreases/increases) uniformly. Subsequently, the velocity becomes (uniform/non-uniform). Finally, the velocity of the object (decreases/increases) uniformly. The object undergoes (zero/uniform/non-uniform) acceleration initially. Next, it has (zero/uniform/non-uniform) acceleration. Finally, it undergoes uniform (deceleration/acceleration).

2.1.3g Quantitive Analysis of the Velocity-Time Chart

If an object is moving with uniform acceleration, a ticker tape can be used to calculate its acceleration. For such an object, the distance between consecutive dots increases uniformly.

A B C

D E F0

Length of tape/cm Length of tape/cm Length of tape/cm

Length of tape/cm Length of tape/cm Length of tape/cm

Number of ticks

Number of ticks

Number of ticks

Number of ticks

Number of ticks

Number of ticks

0 0

000

0.5 cm 1.0 cm 1.5 cm 2.0 cm 2.5 cm


Name: ____________________________________ Form: _________ Date: ___________

Notice that distance between the first two dots is 0.5 cm. Distance between subsequent consecutive dots increases by 0.5 cm.

(i) Calculation of the acceleration directly from the ticker tape The acceleration of an object can be calculated directly from a ticker tape above.

1. Determination of time

Assume the frequency of the ticker timer is 50 Hz. The time for 1 strip, t = 1 tick = 0.02 s

2. Determination of the Initial Velocity u

Initial displacement, s1 = Length of the first strip = 0.5 cm.

3. Determination of the Final Velocity v

Final Displacement, s2 = Length of the fifth strip = 2.5 cm.

4. Calculation of the Acceleration

For an object moving with uniform acceleration,

(ii) Calculation of the acceleration using a ticker tape chartThe acceleration of an object can also be calculated using a ticker tape chart. A ticker tape is first cut up into ten-tick strips. They are then arranged into a chart as shown.


Name: ____________________________________ Form: _________ Date: ___________

1. Determination of time

Assume the frequency of the ticker timer is 50 Hz. The time for 1 tick = 0.02 s. The time for each ten-tick strip, t = 0.02 x 10 = 0.2 s.

2. Determination of the Initial Velocity u

Initial displacement, s1 = Length of the first strip = 30 cm.

3. Determination of the Final Velocity v

Final Displacement, s2 = Length of the fifth strip = 130 cm.

4. Calculation of the Acceleration

For an object moving with uniform acceleration,

2.2 Motion Graphs

2.2.1 Types of Motion Graphs

The three type of motion graphs are: 1. Displacement-time graphs (s-t graphs) 2. Velocity-time graphs (v-t graphs) 3. Acceleration-time graphs (a-t graphs)

Why is the number of strips to determine the acceleration taken as 5 and NOT 6? The time taken is for the velocity to increase from 150 cm s-1 to 650 cm s-1.

How many times does the velocity increase from 150 cm s-1 to 650 cm s-1? 5 NOT 6!

v

t

130 cm

30 cm

Note that the velocity increases 5 times from the first strip to the

sixth strip

I

VI


Name: ____________________________________ Form: _________ Date: ___________

2.2.2 Displacement-Time Graphs (s-t graphs)

1. The velocity of an object can be calculated from the gradient of the displacement-time graph (s-t graph).

∆s = Change in s = s2 – s1 = sΔt = Change in t = t2 – t1 = t

Comparing the equation for the gradient and that for velocity, we can conclude that

Velocity = Gradient of the s-t graph

Examples: What is the velocity of the object in each case?

2. The shape of a displacement-time graph (s-t graph) can tell us when an object is a. at rest, b. moving with uniform velocity, or c. moving with non-uniform velocity (accelerating or decelerating).

3. A horizontal displacement-time graph (s-t graph) tells us that an object is at rest (not moving). o An object that is at rest has zero velocity and zero acceleration.

s/m

0

Δs = s2 – s1

Δt = t2 – t1 (x1, y1)

t/s

(t2, s2)

s/m

0

8

t/s10

(a)

t/s0

5

s/m

20

(b)

t/s0

12

6

s/m(c)

t/s0

4

s/m

16

8

(d)


Name: ____________________________________ Form: _________ Date: ___________

4. A sloped displacement-time graph (s-t graph) tells us that an object is moving with uniform velocity. o An object that is at rest has zero velocity and zero acceleration.

5. A curved displacement-time graph (s-t graph) tells us that an object is moving with non-uniform velocity, that is, the object is undergoing acceleration or deceleration.

Example

Figure 1 shows the displacement (s) – time (t) graph for an object moving in a straight line. Describe the motion of the object for each of the time interval shown. a. AB: Velocity –

Acceleration –

b. BC: Velocity – Acceleration –

c. CD: Velocity – Acceleration –

d. DE: Velocity – Acceleration –

2.2.2 Velocity-Time Graphs (v-t graphs)

1. The acceleration of an object can be calculated from the gradient of the velocity-time graph (v-t graph).

s

t0

Sloped line

s

0 t

Horizontal line

Object is NOT moving

(Velocity = 0)(Acceleration = 0)

s

t0

s

t0Object is moving

with uniform velocity

(Acceleration = 0)

Curved line

Object is moving with non-uniform velocity Object is accelerating

Object is moving with non-uniform velocity

Object is decelerating

Curved line

C Ds

tA

B

E

Figure 1


Name: ____________________________________ Form: _________ Date: ___________

∆y = Change in y = y2 – y1

Δx = Change in x = x2 – x1

Comparing the equation for the gradient and that for velocity, we can conclude that

Velocity = Gradient of the s-t graph

2. The displacement of an object can be calculated from the area under a velocity-time graph (v-t graph).

3. The shape of a velocity-time graph (v-t graph) can tell us when an object is a. at rest, b. moving with uniform velocity, or c. moving with uniform acceleration.

2.2.3 Acceleration-Time Graphs (a-t graphs)

1. Like s-t graphs and v-t graphs, the shape of an acceleration-time graph (a-t graph) can tell us more about the motion of an object.

v/m

0

Δy = y2 – y1

Δx = x2 – x1 (x1, y1)

t/s

(x2, y2)

v

t0

Sloped line

v

0 t

Horizontal line which coincides with the t-axis

Object is NOT moving

(Velocity = 0)(Acceleration = 0)

v

t0

v

t0Object is moving

with uniform velocity

(Acceleration = 0)

Object is moving with increasing velocity

Object is undergoing uniform acceleration

Object is moving with decreasing velocity

Object is undergoing uniform deceleration

Horizontal line Sloped line

If the velocity-time graph is a curved line, what can you tell about the velocity and the acceleration of the object? Is the velocity uniform? Is the acceleration uniform? Curved

line

v

t0

Curved line

v

t0


Name: ____________________________________ Form: _________ Date: ___________

2. We will only look the significance of horizontal graphs here.

Pythagoras Theorem

o Pythagoras Theorem is widely used in vector calculations.

To find the hypotenuse or the other two sides of a triangle, Pythagoras Theorem is used.

OR

o It is useful to remember the ratios for the hypotenuse and the other two sides of a triangle.

a b c a b c3 4 5 5 12 136 8 10 10 24 263x 4x 5x 5x 12x 13x

o It is also useful to remember the equations for sine, cosine and tangent.

(SOH) (CAH) (TOA)

a

t0Horizontal line

a

0 t

Horizontal line which coincides with the t-axis

Object is zero acceleration. We cannot tell if the object is at rest

or in motion unless more information is given.

a

t0

Object is moving with uniform positive acceleration. We cannot tell if the

object is accelerating or decelerating unless more information is given.

If the acceleration-time graph is a sloped line, what can you tell about the velocity and the acceleration of the object?

Object is moving with uniform negative acceleration. We cannot tell if the object

is accelerating or decelerating unless more information is given.

Note: a = oppositeb = adjacentc = hypotenuse

A

BC

b

a

c

θ

2.1 Linear Motion

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