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Be able to define the term 'speed', symbol v.
Be able to name and give the symbols of the SI units for distance, time and speed
Be able to use the relationship v = Δd/Δt to solve a range of problems involving speed, distance and time.
Be able to interpret how an object is moving from the shape of a distance-time graph.
Be able to draw distance-time graphs from the description of the motion of an object.
Be able to draw distance-time graphs from results obtained in motion experiments.
Be able to use the gradient (slope) of a distance-time graph to determine the distanced travelled.
An object is in motion if its location is changing over time.
If an object is not moving it is described as stopped or stationary, and would have a speed of 0 ms-1.
Objects in motion can be described by four different quantities:- distance, time, speed and acceleration.
Distance, symbol d, is measured in metres (m) and is described as how far an object has travelled
Time, symbol t, is measure in seconds (s) and is described as how long a journey has taken.
Speed, symbol v, is measured in meters per second (ms-1) and is described as how fast an object is travelling. Speed is given the same symbol as velocity, v, which is a quantity used in Level 2 & 3 Physics. Speed is what we call a scalar quantity with just a size, e.g. a speed of 5ms-1. Whereas, velocity is what we call a vector quantity with a size and a particular direction, e.g. a velocity of 5 ms-1north.
The last quantity, which we will discuss more later is acceleration. Acceleration, symbol a, is measured in metres per second squared (ms-2) and is described as how an object's speed is changing over time.
Speed is how far an object has travelled in a particular time. We can calculate the speed of an object by dividing distance travelled by time taken.
In the exam you will be given the following equation:
The symbol 'Δ' means a change is something, e.g. change in distance (Δd), or change in time (Δt). The symbol 'Δ' is the upper case Greek letter delta, so we say delta d for Δd, and delta t for Δt.
We can rearrange the equation to find out v, Δd and Δt. See the diagram below.
A car travels 50.0 m in 3.90 s. What is the speed of the car?
Workings:
Δv = d ÷ Δt Δv = 50.0 m ÷ 3.90 s = 12.8 ms-1 (written to 3 significant figures)
Answer: Δv = 12.8 ms-1 (written to 3 significant figures)
A sprinter runs 100 m in 12.5 s. Work out her speed in a) ms-1 and b) kmh-1 (kilometres per hour).
a) Workings:
Δv = d ÷ Δt Δv = 100 m ÷ 12.5 s = 8.00 ms-1 (written to 3 significant figures)
a) Answer:
Δv = 8.00 ms-1 (written to 3 significant figures)
b) Workings:
Convert 8.00 ms-1 to kmh-1 by first dividing by 1000 then multiplying by 3600 (or just multiply by 3.6). There are 1000 m in 1 km, and 60 x60 = 3600 s in 1 hour.
(8.00 ÷ 1000) x 3600 = 28.8 kmh-1 OR 8.00 x 3.6 = 28.8 kmh-1
b) Answer:
Δv = 28.8 kmh-1
A jet can travel at 350 ms-1 . How far will it travel in a) 30 s and b) 5 minutes?
a) Workings:
d = Δv x Δt d = 350 ms-1 x 30 s = 10500 m
a) Answer: d = 10500 m (or 10.5 km)
b) Workings:
First convert 5 minutes to seconds by multiplying by 60
5 minutes = 5 x 60 s = 300 s
d = Δv x Δt d = 350 ms-1 x 300 s = 105000 m
b) Answer: d = 105000 m (or 105 km)
A snail crawls at a speed of 0.0004 ms-1 . How long will it take to climb a garden cane 1.6 m high?
Workings:
Δt = d ÷ Δv Δv = 1.6 m ÷ 0.0004 s = 4000 s
Answer:
Δt = 4000 s
If an object moves in a straight line, the distance travelled can be represented by a distance-time graph.
In a distance-time graph, the gradient (slope) of the line is equal to the speed of the object.
The steeper the line (and the greater the gradient) the faster the object is moving.
A horizontal line on the distance-time graph shows that the object is stationary (not moving).
By calculating the gradient of the distance-time graph you can determine the speed of an object.
To calculate the gradient or slope of the graph you should use:
slope or gradient = change in y axis (rise) = Δd
change in x axis (run) Δt
Be able to define the term 'acceleration', symbol a.
Be able to name and give the symbol of the SI unit for acceleration.
Be able to use the relationship a = Δv/Δt to solve a range of problems involving acceleration, velocity (speed), and time.
Be able to draw speed/time graphs from results obtained in motion experiments.
Be able to interpret how an object's motion from the shape and gradient of a speed-time graph.
Be able to calculate acceleration from the gradient (slope) of speed-time graphs.
Be able to calculate distance travelled by determining the area under a speed-time graph.
Be able to solve problems using acceleration due to gravity where g = 10 ms-2 near the Earth's surface.
Acceleration is the change is speed (change in velocity) of an object.
The object can be accelerating (getting faster) or decelerating (getting slower).
Acceleration, symbol a, is measured in metres per second squared (ms-2) and is described as how an object's speed is changing over time.
The acceleration of an object can be found by using this formula:
acceleration, a = change in speed = Δv
change in time Δt
The change is speed (change in velocity), Δv, is the final speed (final velocity) minus the initial speed (initial velocity)
Δv = vf - vi
Acceleration can be found from a speed-time graph. The instantaneous speed is plotted against the time elapsed.
By calculating the gradient of the spped-time graph you can determine the acceleration of an object.
To calculate the gradient or slope of the graph you should use:
slope or gradient = change in y axis (rise) = Δv
change in x axis (run) Δt
The slope or gradient of the speed-time graph can be used to describe the acceleration.
Slope (gradient) of line Acceleration indicated
upward slope constant acceleration
downward slope constant deceleration
horizontal line constant speed/velocity
steep slope high acceleration
gentle slope low acceleration
A car is travelling at 20 ms-1. It accelerates steadily for 5.0 s, after which time it is travelling at 30 ms-1. What is the acceleration of the car?
Workings:
Δv = final v - initial v = 30 ms-1 - 20 ms-1 = 10 ms-1
a = Δv ÷ Δt a = 10 ms-1 ÷ 5.0 s = 2.0 ms-2 (written to 2 significant figures)
Answer: Δv = 2.0 ms-1 (written to 2 significant figures)
An object strikes the ground travelling at 40 ms-1. It is brought to rest in 0.02 s. What is its acceleration?