Sectieoverzicht
-
-
GPB Education. (2019). What are vectors and scalars? (Standard YouTube licence)
Distance is a scalar quantity representing the interval between two points. It is just the magnitude of the interval. Distance is the total movement of an object without any regard to direction. We can define distance as to how much ground an object has covered despite its starting or ending point.
The SI unit for distance and displacement is m.
Distance here will be = 4m + 3m + 5m = 12 m
Distance Formula: Δd= d1 + d2
Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude. It is represented as an arrow that points from the starting position to the final position.
For example- If an object moves from A position to B, then the object’s position changes. This change in the position of an object is known as displacement.
Look at the picture above, boy travels from D to A, A to B, B to C and C to D. Displacement from D to D (which are our initial and final points) is zero. However, distance traveled is not zero. It is equal to the perimeter of the rectangle.
Images: Harrage, E. (2023). Distance and displacement (CC BY)
John walks from the point A to B to C. What does the distance he travel?
What is the displacement?
Distance travelled: He travels from A to B to C. Distance from A to B is 4 m and B to C is 3 m. Their sum will give us total distance:
4+3=7 m
Then it’s time to calculate displacement. Displacement is a vector quantity. So it must have both magnitude and direction. In this example the initial point is A and the final point is C.
Displacement vector is an internal between the initial and final points. As shown in the diagram the interval between A to C is 5 m. So, the displacement vector is 5 m and the direction is from the point A to C.
Image: LibreText Physics. (2022). Displacement (CC BY 4.0)
A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by Χ . The −4.0 m displacement of the passenger relative to the plane is represented by an arrow toward the rear of the plane. Notice that the arrow representing his displacement is twice as long as the arrow representing the displacement of the professor (he moves twice as far).
Note that displacement has a direction as well as a magnitude. The displacement is 4.0 m toward the rear.
In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction).
The airplane passenger’s initial position is x0=6.0 m and his final position is xf=2.0 m, so his displacement is:
Δx=xf−x0=2.0 m−6.0 m=−4.0 m.
His displacement is negative because his motion is toward the rear of the plane.
Look at the picture given below. An object moves from point A through B, C, D, E and stops at point F.
a) Find final displacement.
b) Find distance taken from point A to D.
We find final displacement by drawing straight line from point A to final point F. As you can see from the graph, object changes its position 8m.
Displacement = Final position - Initial position Displacement = 10m - 2m = 8m
b) We find distance taken by object:
A to B =10 - 2 = 8 m
B to C = 10 - 2 = 8 m
C to D = 10 - 6 = 4 m.
Total distance taken from point A to D is = 8 m + 8 m + 4 m =20 m
Distance Displacement The complete length of the path between any two points is called distance Displacement is the direct length between any two points when measured along the minimum path between them Distance is a scalar quantity as it only depends upon the magnitude and not the direction Displacement is a vector quantity as it depends upon both magnitude and direction Distance can only have positive values Displacement can be positive, negative and even zero
QUICKFACTOpedia. (2018). Distance vs Displacement (Standard YouTube licence)
GPB Education. (2019). What are distance and displacement? (Standard YouTube licence) -
-
-
-
Distance is how far an object moves. It does not include an associated direction, so distance is a scalar quantity.
Speed is the rate of change of distance – it is the distance travelled per unit time. Like distance, speed does not have an associated direction, so it is a scalar quantity.
Typical speeds - When people walk, run, or travel in a car their speed will change. They may speed up, slow down or pause for traffic.
It is not only moving objects whose speed can vary. The speed of the wind and the speed of sound also vary. A typical value for the speed of sound in air is about 330 m/s.
To work out an object’s speed you need to know the distance it has travelled, and the time taken.
Calculate speed using the equation: Speed = distance/time.
Average speed is calculated as the distance traveled over the total time of travel.
For example:
If you by car travelled from Paje to Potoa, as distance of 72.5 km, in 1 hour, your average speed would be:
Average speed = 72.5/1 = 72.5 km/hr.
That's not to say that at every point of the journey, you will be travelling at 72.5 km/hr. Some of the time you will be travelling faster and at other times slower.
The velocity of an object is its speed in a particular direction. Velocity is a vector quantity because it has both a magnitude and an associated direction. To calculate velocity, displacement is used in calculations, rather than distance.
Unlike distance, which is a scalar quantity, displacement is a vector quantity.
It includes:
- the distance travelled, measured in a straight line from start to finish
- the direction of the straight line
Images: Harrage, E. (2023). Speed and velocity (CC BY)
Average velocity is defined as the change in position (or displacement) over the total time of travel.
Image: CourseHero. Openstax. (nd). Average velocity (CC BY-SA)
For example:
Average Speed vs. Average Velocity: If you started walking from one corner and went all the way around the rectangle in 30 seconds, your average speed would be 0.47 m/s, but your average velocity would be 0 m/s.
Example:
During a 4 s time interval, the runner’s position changes from x1 = 60 m to x2= 40 m. What is the runner’s average velocity?
Solution:
The change in the displacement of the runner is calculated as follows:
Δx = x2 – x1 = 40 m – 60 m = -20 cm
Δt = 4 s
vaverage = Δx/Δt = -20 m/ 4 s = -5 m/s
GPB Education. (2019). What are speed and velocity? (Standard YouTube licence) -
-
-
-
Acceleration is the change in speed or velocity of an object over a certain time. It can be calculated by dividing the change in velocity by the total time.
Acceleration is the rate of change in speed (or velocity).
It is defined as follows:
In physics, the following standard symbols are used to represent the quantities shown in the acceleration equation:
u – initial speed
v – final speed
a – acceleration
t - time
to give the final speed of an object after it has accelerated.
Acceleration is measured in metres per second per second ( m.s-2 ), often pronounced as 'metres per second squared'.
Sometimes when we are describing motion we use the term 'constant speed'.
An object is travelling at a constant speed when its instantaneous speed has the same value throughout its journey. For example, if a car is travelling at a constant speed the reading on the car's speedometer does not change. The speedometer displays the speed of the car at each moment in time throughout the car's journey. In cases like this, where the motion involves only constant speed, the instantaneous speed of the object can be worked out using the relationship:
Speed = distance travelled/time taken
We also sometimes refer to a moving object as having a 'constant acceleration' or a 'uniform acceleration'. A constant or uniform acceleration means that the speed of the object changes by the same amount every second.
When the speed of an object is decreasing with time (i.e. slowing down), the object's speed is changing and so, by definition, the object is accelerating. If the velocity of an object is decreasing, the acceleration calculated will be negative.
This is sometimes known as a deceleration.
Example question:
Find the acceleration of a car, if it starts at 10 m/s and it reaches 30 m/s in 4 seconds.
The change in velocity is v – u, which is 30 – 10 = 20 m/s.
The acceleration is the change in velocity ÷ time, which is 20 m/s ÷ 4 s = 5 m/s2.
What happens if the car slows down? Can you work out the acceleration now?Find the acceleration of the car, if it starts at 20 m/s and it reaches 12 m/s in 2 seconds.
The change in velocity is v – u, which is 12 – 20 = -8 m/s.
The acceleration is the change in velocity ÷ time, which is -8 m/s ÷ 2 s = -4 m/s2.
A minus sign means that the car is decelerating.
Acceleration and velocity
Acceleration is the rate of change of velocity. It is the amount that velocity changes per unit time.
The change in velocity can be calculated using the equation:
Change in velocity = final velocity - initial velocity
The average acceleration of an object can be calculated using the equation:
This is when:
- acceleration (α) is measured in metres per second squared (m/s²)
- change in velocity (∆v) is measured in metres per second (m/s)
- time taken (t) is measured in seconds (s)
- If an object is slowing down, it is decelerating (and its acceleration has a negative value). This is also known as retardation.
Example question:
A car takes 8.0 s to accelerate from rest to 28 m/s. Calculate the average acceleration of the car.
final velocity, v = 28 m/s
initial velocity, u = 0 m/s (because it was at rest - not moving)
change in velocity, ∆v = (28 - 0) = 28 m/s
α = Δv/t
So: 28/8 = 3.5 m/s2
FuseSchool. (2020). Acceleration (CC BY) -
-
-
-
The first graph in the diagram above shows a velocity-time graph for a car travelling at a constant velocity. It is not accelerating or decelerating.
As shown in the second the acceleration is gently increased so the graph slopes upwards.
In the third graph the car accelerates more rapidly, and since its velocity would increase more rapidly the graph would be steeper.
The acceleration is shown by the slope (gradient) of the velocity-time graph.
Here the velocity-time graph of a car starting off at one set of traffic lights and stopping at the next set of lights. Can you see that the car accelerates from A to C, travels at a constant velocity between C and D and then decelerates to stop at E?
Between which two points is the motor bike accelerating most rapidly? Which part of the graph might show the motorbike travelling at a constant speed?
When does the rider start braking? Which part of the graph shows the rider hitting a brick wall?
Here is a velocity-time graph of a lift climbing from the ground floor to the top of the building.
At which point (A, B or C) is it on the ground? A
At which point is it travelling the fastest? B
What is its maximum velocity? 10 m/s
How long did it take to reach this velocity? 5 seconds
It decelerates and stops at the top of the building at C. How long did the journey take? 7 seconds
We can use the slope to find its acceleration:
acceleration = change in velocity/ time taken for the change
acceleration = 10 m/s ÷ 5 s
acceleration = 2 m/s2
Look at each of the lettered sections OA, AB, BC, CD, DE and EF on the speed-time graph.
Images: Harrage, E. (2023). Velocity-time graphs. (CC BY)
a) Choose the correct word from the list below to describe the motion taking place in each section.
constant speed
stationary
(i) OA (ii) AB (iii) BC (iv) CD (v) DE (vi) EF
b) At what time during the journey did the bus reach its greatest speed?
c) How long did the bus stop for during its journey?
d) During which section of the journey did the bus have the greatest acceleration?
e) Calculate the acceleration of the bus during section DE.
f) Which section could be described as having a negative acceleration?
Answers:
a) OA - the speed of the bus is changing. Therefore, according to the definition, the bus has an acceleration.
AB - the speed does not change. The bus has a constant speed of 10 m/s.
BC - the bus is slowing down. Its speed is changing. Therefore, according to the definition, the bus has an acceleration.
CD - the speed of the bus is 0 m/s over this section. The bus is stationary.
DE - the speed of the bus is changing. The bus has an acceleration.
EF - the speed does not change. The bus has a constant speed of 5 m/s.
b) After 5 s.
c) 3 s.
d) BC - this part of the graph has the steepest slope.
e) To find the acceleration in section DE use the formula
FuseSchool. (2020). Velocity-time graphs (Standard YouTube licence) -
-
-
-
What is uniform motion?
This type of motion is defined as the motion of an object in which the object travels in a straight line and its velocity remains constant along that line as it covers equal distances in equal intervals of time, irrespective of the duration of the time.
What is uniform acceleration?
When an object is travelling in a straight line with an increase in velocity at equal intervals of time, then the object is said to be in uniform acceleration. Free falling of an object is an example of uniform acceleration.
There are 4 equations which can be used whenever an object travels with constant, uniform acceleration in a straight line.
These equations use 5 symbols: s, u, v, a, t.
s = distance travelled (meters)
u = initial velocity (m/s)
v = final velocity (m/s2)
a = acceleration (m/s2)
t = time taken (seconds)
Suppose an object is travelling at a velocity u and then moves with a uniform acceleration a for a time t. Its velocity is then v and it has travelled a distance s.
Image: Harrage, E. (2023). Equations of motion (CC BY)
You can remember the 5 symbols by "suvat". If you know any three of suvat, the other two can be found.
In questions, initially at rest means u = 0. A negative number for the acceleration means the object is decelerating (slowing down).
Example:
A cheetah starts from rest and accelerates at 2 m/s2 for 10 seconds. Calculate:
a) the final velocity
b) the distance travelled.
First write out suvat:
s = ?
u = 0
v = ?
a = 2 m/s2
t = 10 s
a) Use equation 1: v = u + at
v = 0 + 2 x 10
v = 20 m/s
b) Use equation 2:
Example:
A nail is fired from a nail gun into a fixed block of wood. The nail has a speed of 380 m/s just as it enters the wood. The nail comes to rest after penetrating 60 mm into the wood. Find the time taken for the nail to come to rest. Assume that the retarding force on the nail is constant as it penetrates the wood.
S = 60 mm or 60 × 10-3 cm
u = 380 m/s
v = 0
a = ?
t = ?
Use the equation: v2 = u2 + 2as
02 = 3802 + 2a × 60 × 10-3
a = -1,203,333 ms-2
v = u +at
0 = 380 + (-1,203,333)t
t = 0.0003157 seconds
The time taken for the nail to come to rest is 3.16 × 10-4 s.
Myhometuition. (2014). Equations of uniform acceleration (Standard YouTube licence) -
Leerlingen moetenEen cijfer behalenHet slaagcijfer behalen
-
-
-
The force of gravity pulls on all objects on Earth. If all objects are allowed to fall, they will accelerate downwards. If there is no air resistance, then all objects would fall at the same rate.
Acceleration due to gravity = 9.8 m/s2. For simplicity 10 m/s2 is usually used.
This means that for an object falling with no air resistance, the velocity after 1 second is 10 m/s, after 2 seconds velocity is 20 m/s and so on.
Image: IFP news. (2015). Available online here. CC BY 4.0)
In practice, there usually is air resistance. If the skydiver in the photo falls a long way without a parachute, then because of friction she reaches a final or terminal velocity of about 50 m/s, which is the same speed as a fast racing car.
Raindrops, snowflakes and seeds for example, all fall at their own terminal velocity.
A parachute is designed to make the air resistance as large as possible. With a parachute the terminal velocity is 8 m/s.
Image: Pxhere (CC0 PD)
At terminal velocity, the forces on the object are balanced:
Force of gravity (weight) = force of air resistance
(downwards) (upwards)
This is an example of Newton’s first law, because there is no resultant force on the object, it continues to move at a constant speed in a straight line.
Example:
A ball is thrown vertically upwards at 20 m/s. Ignoring air resistance and taking gravity = 10 m/s2, calculate:
a) How high it goes.
b) The time taken to reach this height.
c) The time taken to return to its starting point.
When travelling upwards it is decelerating so a = - 10 m/s2.
At the moment when it reaches its highest point, v = 0.
Write:
s = ?
u = 20 m/s
v = 0
a = - 10 m/s2
t = ?
a) v2 = u2 + 2as
0 = 202 + 2(-10)s
20s = 202
s = 20 m
b) v = u + at
0 = 20 + (-10) t
t = 2 seconds
c) It also takes 2 seconds to fall down again, so in time take to return to its starting point = 4 seconds.
Exam Solutions. (2014). Vertical motion under gravity (Standard YouTube licence)
-
-
