VECTOR & SCALAR QUANTITY, DISTANCE/DISPLACEMENT, SPEED/VELOCITY, ACCELERATION, DISTANCE/DISPLACEMENT –TIME GRAPH, SPEED/VELOCITY–TIME GRAPH
CONTENT:
Scalar & Vector Quantity
Distance & Displacement
Speed & Velocity
Acceleration & Retardation
Distance/Displacement - Time Graph
Speed/Velocity - Time Graph
https://youtu.be/Ix3htsNmSZY
SCALAR & VECTOR QUANTITY
A
scalar quantity is defined as a quantity that has magnitude only but no direction. Typical examples of scalar quantities are time, distance, speed, temperature, volume, work, power, electric potential etc. A scalar quantity or parameter has no directional component, only magnitude. For example, the units for time (minutes, days, hours, etc.) represent an amount of time only and tell nothing of direction. Additional examples of scalar quantities are density, mass, and energy.
A
vector quantity is defined as a quantity that has both magnitude and direction. Typical examples of vector quantities are velocity, displacement, acceleration, force, momentum, moment, electric field intensity etc. To work with vector quantities, one must know the method for representing these quantities. Magnitude, or “size” of a vector, is also referred to as the vector's "displacement." It can be thought of as the scalar portion of the vector and is represented by the length of the vector. By definition, a vector has both magnitude and direction. Direction indicates how the vector is oriented relative to some reference axis, as shown in Figure 1. Using north/south and east/west reference axes, vector "A" is oriented in the NE quadrant with a direction of 45 north of the o EW axis. Giving direction to scalar “A" makes it a vector. The length of “A” is representative of its magnitude or displacement.
https://youtu.be/Pj8Zh0A-uLU
https://youtu.be/iLB_4Wu2QOg
Distance & Displacement
Distance is the gap or space between two points, measured in meters (m).It is a scalar quantity & can be calculated mathematically as;
Distance = Average speed x time taken ……………………………………..1.
Displacement is distance in a specified direction, measured in meters (m). It is a vector quantity & can be calculated mathematically as;
Displacement = Average x time taken ………………………………………..2.
Speed & velocity
Speed :the rate at which a body covers a distance is called the speed of the body. Thus
Speed = distance/Time ………………………………….3.
The S.1 unit of speed is meter per second (ms-1)
When a body covers equal distance in equal time intervals, no matter how small the time interval may be, it is said to be a uniform speed or constant speed
Velocity is defined as the rate of change of displacement with time. It is also the rate of change of distance in a specified direction. The S.1 unit of velocity is meter per second (ms-1). In everyday language, the words velocity and speed are often used interchangeably. In the study of motion it is necessary to distinguish between the two. In speed, no direction is specified, but in velocity it is necessary to specify direction.
When a body moves with equal displacement in equal time interval, no matter how small the time interval may be, the velocity is said to be a uniform velocity or constant velocity.
Velocity = Displacement/Time……………………………………….4
Acceleration & Retardation
Acceleration is defined as the increasing rate of change of velocity. It is measured in m/s2.
Acceleration (a) = Increasing Velocity change/Time taken ……………………………………5.
When the velocity of a moving body increases by equal amount in equal intervals of time, no matter how small the time intervals may be, it is said to move with uniform acceleration.
Retardation is defined as the decreasing rate of change of velocity. It is measured in m/s2.It is also known as deceleration or negative acceleration
Retardation (ar) = Decreasing Velocity Change/Time Taken ………………………………6.
https://youtu.be/un2BJgVL2-I
EQUATION OF UNIFORMLY ACCELERATED MOTION
S = (v+u)/2t ………………………………………………………7
v = u + at ……………………………………………………….8
v² = u²+ 2aS ……………………………………………………….9
S = ut + ½ at² ……………………………………………………….10
Equations (7) to (10) are called
equations of uniformly accelerated motion and could be used to solve problems associated with uniformly accelerated motion
where u- initial velocity( m/s), v – final velocity (m/s), a – acceleration (m/s2), s – distance covered and t – time (m)
Example - A car moves from rest with an acceleration of 0.2mls2 . Find its velocity when it has moved a distance of 50m
Solution:
a = 0.2mls² , S = 50m, u = 0m/s , v = ?
v² = u² + 2 as
v² = 0² + (2x0.2x50) = 20
v = √20 m/s
https://youtu.be/UYS5fFMd8cM
EVALUATION
1. State the differences & similarity between speed & velocity 2.A car has a uniform velocity of 108km/hr. How far does it travel in ½ minute?
Distance/Displacement- Time Graph
If the body is moving with uniform velocity, the velocity is equal to the slope or gradient of the straight line graph.
If the velocity is non – uniform, the velocity at a point is the gradient or slope of the tangent at that point.
Speed/Velocity - Time Graph
Speed/velocity–graph is more useful than distance/displacement-time graph because it gives additional information about the motion performed by the body. Such information include acceleration, retardation, total distance covered & average speed or velocity
Example - A car starts from rest and accelerates uniformly until it reaches a velocity of 30mls after 5 seconds. It travels with uniform velocity for 15 seconds and is then brought to rest in 10s with a uniform retardation. Determine (a) the acceleration of the car (b) The retardation (c) The distance covered after 5s (d) The total distance covered (use both graphical and analytical method)
The velocity – time diagram for the journey is shown above, from this diagram
b. the retardation = slope of BC = CB / CD
= (0-30) / (30-20) = -30/10
= -3mls2 (the negative sign indicate that the body is retarding)
c. Distance traveled after 5s = area of A E O
= ½ x b x h
= ½ x 5 x 30
= 75m
d. Total distance covered = area of the trapezium OABC
= ½ (AB + OC) AE
= ½ (15 + 30) 30
= 675m.
Using equations of motion.
a) U = O, V = 3, t = 5
V = u + t
a = v-u/t = 30 – 0 / 5
a = 30/5 = 6ms-2
b) a o in
a = v – u / t = 0-30 / 10
a = -3 mls2
(c) S = ( u + v) 5
2
= 30 / 2 x 5
= 75m
(d) To determine the total distance travelled we need to find the various distance for the three stages of the journey and then add them.
for the 1st part S= 75m from (c)
for the 2nd stage where it moves with uniform velocity.
S = vt
= 30 x 15
= 450m
for the last stage S = ½ (u + v) t
= ½ (30 + 0) 10
= 150m.
Total distance = 75 + 450 + 100 = 675m.
EVALUATION
1. A train slows from 108km/hr with uniform retardation of 5mls2. How long will it take to reach 18km/hr and what is the distance covered.
2. Why is velocity – time more useful than displacement time graph?
Reading Assignment -
www.google.com (click on google search, type “ distance & displacement ”, click on search) & New school physics by M.W.Anyakoha,Ph D Pg 14 – 18
ASSIGNMENT
1. A body accelerates uniformly from rest at the rate of 3ms-2 for 8 seconds. Calculate the distance covered. (a) 12m (b) 24m (c) 48m (d) 96m
2 A particle starts from rest and moves with a constant acceleration of 0.5 ms-2 . The distance covered by the particle in 105s is (a) 2.5m (b) 5.0m (c) 25.0m (d) 50.0m
3.A car moving with speed 90kmh-1 was brought uniformly to rest by the application of brakes in 10s. How far did the car travel after brakes were applied (a)120m(b) 150m (c) 125m(d)15km
4. A body starts from rest and accelerates uniformly at 5mls2 until it attains a velocity of 25mls. Calculate the time taken to attain this velocity. (A) 2.5s (b) 5.0s (c) 10s (d) 125s
5. The distance traveled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resulting graph is linear. The slope of this graph is a measure of (a) initial displacement (b) initial velocity (c) acceleration (d) half of acceleration
THEORY
1. When is a body said to be moving with uniform speed?
2. A motor car accelerates uniformly from rest at 5mls2 until it reaches a speed of 20ms-1.zbdgv It travels at this speed for 4 seconds. The brakes are then applied and the car comes to rest with uniform retardation in a further 8 seconds. Draw a sketch showing a graph of speed against time. Use your graph to determine
(a)How far the car travels after the brakes are applied?
(b)The time during which the car accelerated
(c)Total distance covered
(d) Average speed
Concept of distance, speed, velocity and uniform speed/velocity
https://youtu.be/Jyiw6KkedDY
i. DISTANCE: This is the separation or space between two points. It is measured in meters and it is a scalar quantity.
ii. DISPLACEMENT: It is distance in a specified direction. It is a vector quantity and it is measured in meters.
iii. SPEED: It is the rate of change of distance moved with time. The unit is m/s and it is a scalar quantity.
Speed = distance/time
(a) UNIFORM SPEED: It is obtained if the rate of change of distance with time is constant or when a body travels equal distances in equal time intervals.
(b) AVERAGE SPEED: If the speed is not constant, The average speed is taken. Average speed is the total distance travelled divided by the total time taken.
Average speed = (Total distance covered)/(Total time taken)
(c) INSTANTANEOUS SPEED: It is the actual speed of a body at any instant.
iv. VELOCITY: It is the rate of change of displacement with time. The unit is m/s. It is a vector quantity.
Velocity = displacement/time
UNIFORM VELOCITY: It occurs when the rate of change of displacement with time is constant or when a body travels equal displacement in equal time interval.
https://youtu.be/Xo3KBoEMDEo
https://youtu.be/QaU9jMHh7gE
EVALUATION:
1. Define speed, velocity and uniform velocity.
2. Differentiate between velocity and speed.
Calculations on speed and velocity
https://youtu.be/EGqpLug-sDk
1. A car covers a distance of 60km in half an hour. What is the average speed of the car in
(a). km/hr (b) m/s
Solution:
(a) time = ½ hour = 0.5 hour
Average speed = (Total distance covered)/(Total time taken) = 60/0.5 = 120km/h
(b) convert km/hr to m/s
1 km/h = 1000/(60×60) m/s
1km = 1000m
1hr = 3600s
120km/h = 120×1000/(60×60) = 3.33m/s
2. A car travelled to Lagos a distance of 150m in 100 seconds. Calculate his average speed.
Average speed = (Total distance covered)/(Total time taken)=150/100=1.5m/s
3. A car covers 1500m in 10 secs. What is the speed in km/hr?
Speed = distance/time=1500/10=150m/s
Convert to km/hr
1 km/h = 1000/(60×60) m/s
1 m/s =(60×60)/1000 km/h
150m/s = (150×60×60)/1000=540km/h
https://youtu.be/P0UYC8S4kUI
EVALUATION:
1. Convert 144km/h to m/s.
2. A car covers a distance of 40m in 2 sec. What is his speed in km/h?
Distance-time graph, Displacement-time graph
If the body is moving with uniform velocity, the velocity is equal to the slope or gradient of the straight line graph.
If the velocity is non – uniform, the velocity at a point is the gradient or slope of the tangent at that point.
TRAVEL GRAPH: It is the graphical representation of the motion of a body. There are Distance-time graph, Displacement-time graph and Velocity-time graph.
(a) DISTANCE-TIME GRAPH FOR UNIFORM MOTION
(b) DISTANCE-TIME GRAPH FOR NON UNIFORM MOTION
(c) DISPLACEMENT-TIME GRAPH FOR UNIFORM VELOCITY
(d) DISPLACEMENT-TIME GRAPH FOR NON UNIFORM VELOCITY
Position, Distance and Displacement
1. Concept of position, Concept of distance and displacement
2. Distinction between distance and displacement.
CONCEPT OF POSITION
The position of an object is its location in space. It is usually expressed in relation to a reference point. To locate an object in space, a co-ordinate system is needed. It is usually a mathematical construct with co-ordinates.
A coordinate system could be two-dimensional as in P(x,y) or three dimensional as in P(x,y,z).
CONCEPT OF DISTANCE & MEASUREMENT
Distance can be define as a physical measurement of length between two points. It does not take into consideration the direction between the two points it measures; hence, it is a scalar quantity. This therefore means that distance has only magnitude but no direction. E.g, 10km.
Distance could be measured using instruments like measuring tape, ruler, venier calliper, micrometer screw gauge, etc.
CONCEPT OF DISPLACEMENT
Displacement is defined as the distance travelled or moved in a specific direction. It takes into consideration the direction between the different points it seeks to measure; hence, displacement is a vector quantity. Thus, it has both magnitude and direction. E.g, 10km due east. The ‘10km’ is the magnitude (or value), while ‘due east’ is the direction.
Both distance and displacement have the same S.I. unit, metre (m). They could also be expressed in kilometre (km), miles, etc.
https://youtu.be/o1NEJRf2zng
Evaluation:
1. Define distance.
2. What is displacement?
3. State the SI unit of distance.
Distinction between distance and displacement
We need to understand the concepts of distance and displacement. Distance is the gap between two points with no regard to direction. On the other hand, displacement is distance covered in a particular direction. Therefore distance is a scalar quantity while displacement is a vector quantity. The only similarity between distance and displacement is that they have the same unit. Let us consider a girl who walked and covered a distance of 20m between two points A and B as shown in fig 1 and fig 2 below
The two activities of the girl are not exactly the same. In both figs. 1 and 2, she covered a distance of 20m. If we are only interested in the distance covered, we can conclude that she did the same thing in fig. 1 and 2 i.e she covered the same distance (20m). If we are interested in both distance and direction, then her displacement in fig. 1 and 2 are not the same. In fig.1 she covered a distance of 20m due east while in fig.2, she covered a distance of 20m due west. From these, we see that distance is a scalar quantity because it has magnitude only while displacement is a vector quantity because it has both magnitude and direction.
https://youtu.be/9z-EIcdJ9VY
Speed/Velocity - Time Graph
Speed/velocity–graph is more useful than distance/displacement-time graph because it gives additional information about the motion performed by the body. Such information include acceleration, retardation, total distance covered & average speed or velocity
Example - A car starts from rest and accelerates uniformly until it reaches a velocity of 30mls after 5 seconds. It travels with uniform velocity for 15 seconds and is then brought to rest in 10s with a uniform retardation. Determine (a) the acceleration of the car (b) The retardation (c) The distance covered after 5s (d) The total distance covered (use both graphical and analytical method)
The velocity – time diagram for the journey is shown above, from this diagram
a. the acceleration = slope of OA
= AE / EO
= (30-0) /(5-0)=30/5
= 6mls2
b. the retardation = slope of BC = CB / CD
= (0-30) / (30-20) = -30/10
= -3mls2 (the negative sign indicate that the body is retarding)
c. Distance traveled after 5s = area of A E O
= ½ x b x h
= ½ x 5 x 30
= 75m
d. Total distance covered = area of the trapezium OABC
= ½ (AB + OC) AE
= ½ (15 + 30) 30
= 675m.
Using equations of motion.
a) U = O, V = 3, t = 5
V = u + t
a = v-u/t = 30 – 0 / 5
a = 30/5 = 6ms-2
b) a o in
a = v – u / t = 0-30 / 10
a = -3 mls2
(c) S = ( u + v) 5
2
= 30 / 2 x 5
= 75m
(d) To determine the total distance travelled we need to find the various distance for the three stages of the journey and then add them.
for the 1st part S= 75m from (c)
for the 2nd stage where it moves with uniform velocity.
S = vt
= 30 x 15
= 450m
for the last stage S = ½ (u + v) t
= ½ (30 + 0) 10
= 150m.
Total distance = 75 + 450 + 100 = 675m.
Using equations of motion.
a) U = O, V = 3, t = 5
V = u + t
a = v-u/t = 30 – 0 / 5
a = 30/5 = 6ms-2
b) a o in
a = v – u / t = 0-30 / 10
a = -3 mls2
(c) S = ( u + v) 5
2
= 30 / 2 x 5
= 75m
(d) To determine the total distance travelled we need to find the various distance for the three stages of the journey and then add them.
for the 1st part S= 75m from (c)
for the 2nd stage where it moves with uniform velocity.
S = vt
= 30 x 15
= 450m
for the last stage S = ½ (u + v) t
= ½ (30 + 0) 10
= 150m.
Total distance = 75 + 450 + 100 = 675m.
https://youtu.be/TG2Y2MDx-zE
EVALUATION
1. A train slows from 108km/hr with uniform retardation of 5mls2. How long will it take to reach 18km/hr and what is the distance covered.
2. Why is velocity – time more useful than displacement time graph?
3. A boy moved continuously for 40secs and covered the following distances in the times stated below:
Distance (m) 200 400 600 800
Time (second) 10 20 30 40
i. Draw the distance-time graph and calculate the speed.
ii. state whether or not the speed is uniform. Give reason(s) for your answer
4. A body accelerates uniformly from rest at the rate of 3ms-2 for 8 seconds. Calculate the distance covered. (a) 12m (b) 24m (c) 48m (d) 96m
5. A particle starts from rest and moves with a constant acceleration of 0.5 ms-2 . The distance covered by the particle in 105s is (a) 2.5m (b) 5.0m (c) 25.0m (d) 50.0m
6.A car moving with speed 90kmh-1 was brought uniformly to rest by the application of brakes in 10s. How far did the car travel after brakes were applied (a)120m(b) 150m (c) 125m(d)15km
7. A body starts from rest and accelerates uniformly at 5mls2 until it attains a velocity of 25mls. Calculate the time taken to attain this velocity. (A) 2.5s (b) 5.0s (c) 10s (d) 125s
8. The distance traveled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resulting graph is linear. The slope of this graph is a measure of (a) initial displacement (b) initial velocity (c) acceleration (d) half of acceleration
ASSIGNMENT
1. A body accelerates uniformly from rest at the rate of 3ms-2 for 8 seconds. Calculate the distance covered. (a) 12m (b) 24m (c) 48m (d) 96m
2 A particle starts from rest and moves with a constant acceleration of 0.5 ms-2 . The distance covered by the particle in 105s is (a) 2.5m (b) 5.0m (c) 25.0m (d) 50.0m
3.A car moving with speed 90kmh-1 was brought uniformly to rest by the application of brakes in 10s. How far did the car travel after brakes were applied (a)120m(b) 150m (c) 125m(d)15km
4. A body starts from rest and accelerates uniformly at 5mls2 until it attains a velocity of 25mls. Calculate the time taken to attain this velocity. (A) 2.5s (b) 5.0s (c) 10s (d) 125s
5. The distance traveled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resulting graph is linear. The slope of this graph is a measure of (a) initial displacement (b) initial velocity (c) acceleration (d) half of acceleration
6. During the same time interval, it is observed that a train travels the same distance as
does a lorry. The two vehicles therefore have the same
A. uniform acceleration B. instantaneous velocity C. initial velocity D. average speed.
7. The time rate of change of displacement is known as
A. speed. B. velocity C. impulse. D. acceleration.
8. The slope of a straight line displacement-time graph indicates the
A. distance traveled B. uniform velocity
C. uniform acceleration D. acceleration at an instant.
9. A car moves with a speed of 30m/s. Calculate the distance travelled in 30s.
A. 30m B. 60 C. 450m D. 900m
10. The speed of an object in rectilinear motion can be determined from the
A. Area under a velocity-time graph. B. Area under a distance-time graph.
C. Slope of a distance-time graph D. Slope of a velocity-time graph.
THEORY
1. When is a body said to be moving with uniform speed?
2. A motor car accelerates uniformly from rest at 5mls2 until it reaches a speed of 20ms-1.zbdgv It travels at this speed for 4 seconds. The brakes are then applied and the car comes to rest with uniform retardation in a further 8 seconds. Draw a sketch showing a graph of speed against time. Use your graph to determine
(a)How far the car travels after the brakes are applied?
(b)The time during which the car accelerated
(c)Total distance covered
(d) Average speed
3. Using a suitable diagram, explain how the following can be obtained from a Distance-time graph or Displacement-time graph.
i. Speed ii. Velocity iii. Instantaneous speed
4. A Car is travelling with a uniform velocity of 72km/h. What distance does he cover in 20s?
5. A car travels with a constant velocity of 45km/h for 10s. What distance does it cover in this time?
READING ASSIGNMENT:
New school physics by M.W. Anyakoha,Ph D Pg 14 – 18
Evaluation:
1. Differentiate distance from displacement in two ways.
2. Why is 5km due east a displacement?
GENERAL EVALUATION:
1. Discuss the concept of distance and displacement.
2. Enumerate the measuring devices for distance.
ASSIGNMENT:
1. which of the following is displacement?
A. 25cm
B. 43inches
C. 52mm due south
D. 88km
2. One of the following is not a measuring device for distance.
A. Vernier calliper
B. Micrometer screw gauge
C. Rule
D. Spring balance
3. A position in two dimensional co-ordinate system has a value P(-2,5). What is the x –coordinate of the point?
A. 5 unit
B. -2 unit
C. 3 unit
D. 7 unit
Essay:
Explain the distinction between distance and displacement.
READING ASSIGNMENT
Read up the topic: ‘’Motion and types of motion’’ in the following text books.
i. Senior Secondary School Physics by P.N. Okeke et al.
ii. New School Physics for Senior Secondary Schools by Anyakoha, M.W.