KINEMATICS DEFINITIONS Distance:
It is the length between two points.
Displacement:
It is the shortest distance between two points with known direction.
Speed:
It is the distance covered per unit time.
Velocity:
It is displacement per unit time. OR It is speed in a given direction.
Average speed/velocity:
It is the total distance/displacement covered over total time.
Uniform speed/velocity:
It is the same distance/displacement covered over every second.
Acceleration:
It is the rate of change of velocity. (Do not say speed in this case as speed is a scalar)
Uniform acceleration:
It is the same velocity change every second. (Do not say constant acceleration or acceleration is same)
FORMULAE 1.
= =
(Unit of speed: m/s or ms-1)
=
2.
(Unit of average speed: m/s or ms-1)
=
3.
=
(Unit of acceleration: m/s2or ms-2)
These are the only formulae required in your syllabus. Equations of motion such as v2 = u2 = 2as are out of syllabus.
MOTION GRAPHS You need to recognize the shape of each graph and be able to distinguish different types of motion from a combination of two or more of these graphs. When describing a v/t graph always refer to the acceleration of the object unless the examiner asks you to describe the velocity of the object.
VELOCITY / TIME GRAPHS Zero acceleration (uniform / constant velocity)
Uniform / constant acceleration (increasing velocity)
velocity (m/s)
velocity (m/s)
time (s)
time (s)
Increasing acceleration (increasing velocity)
Decreasing acceleration ( increasing velocity)
Velocity (m/s)
velocity (m/s)
time (s)
time (s)
Uniform deceleration (decreasing velocity)
Increasing deceleration (decreasing velocity)
velocity (m/s)
velocity (m/s)
time (s)
time (s)
Decreasing deceleration (decreasing velocity) velocity (m/s)
time (s)
Non uniform acceleration
Non uniform deceleration
velocity (m/s)
velocity (m/s)
time (s)
(mixture of increasing and decreasing acceleration)
time (s)
(mixture of increasing and decreasing deceleration)
NOTE: 1.
Gradient of a v/t graph = acceleration.
You can find the acceleration from the graph by finding its gradient as shown: v (m/s)
=
30
=
0
10
2.
t (s)
(acceleration can be found by formula)
= 3 m/s2
=
=
= 3 m/s2
so the acceleration = 3 m/s2
Area under a v/t graph = total distance travelled.
You should be able to find the areas of triangles, trapeziums and rectangles. Let’s see some examples.
v (m/s) 30
=
ℎ
=
10
ℎ
30
= 150. So the distance is 150 m. 0 10
time (s)
v (m/s) 30
0
2
7
9
time (s)
=
ℎ
ℎ
=
30
( 9 + 5)
= 210. So the distance is 210 m.
v (m/s) 10
=
ℎ
ℎ
= 5 x 10 0 5
time (s)
= 50. So the distance is 50 m.
Sometimes you may have to estimate the areas as shown: v Make a triangle out of it and estimate its area.
v
t
t
This is the correct way of drawing a triangle. Wrong ways of drawing triangles: v
v
t
t
DISTANCE / TIME GRAPHS Zero velocity (at rest)
Uniform velocity
distance (m)
distance (m)
time (s)
Increasing velocity
Decreasing velocity
time (s)
NOTE:
time (s)
time (s)
Gradient of s/t graph = velocity.
Always describe acceleration from a velocity/time graph and velocity from a distance / time graph. Do not try to describe acceleration from a distance / time graph.
ACCELERATION / TIME GRAPHS.
You can always derive the a/t graphs from v/t graphs. First deduce the type of motion from the v/t graph and then determine the type of a/t graph you want to make. Zero acceleration
Uniform acceleration
acceleration (m/s2)
acceleration (m/s2)
time (s)
time (s)
Uniform deceleration
acceleration (m/s2)
time (s)
Deducing an acceleration / time graph from a velocity time graph velocity (m/s)
acceleration time (s) time (s)
NOTE: The questions that are set in the exam are either individual graphs or a combination of some of the graphs sketched above. You should be able to determine the type of motion and work out the relevant information from it. When describing motion you can use the word speed or velocity it does not matter. Example: Graph of a car’s journey.
a) b) c) d)
velocity (m/s)
25
describe the motion of the car. find the acceleration of the car. find the deceleration of the car. find the total distance covered by the car.
time (s) 2
5
6
Solution:
a) The car starts off from rest with a uniform acceleration to a maximum speed of 25 m/s in 2 s. Then it continues at this speed for the next 3s. The car decelerates uniformly to rest in the last 1 s. (If it was a distance / time graph then always refer to speed/velocity not acceleration).
b) a = gradient so = =
= 12.5 so the acceleration = 12.5 m/s2.
c) Deceleration has no formula so calculate the acceleration and then make it positive. Data: v = 0 u = 25 m/s t = 1 s. = =
= - 25 m/s2. So deceleration is 25 m/s2.
REMEMBER: deceleration (retardation) is always positive. Acceleration can be negative when something is slowing down and its speed decreases. The value of acceleration in this case will be negative. To describe negative acceleration we use the word deceleration which means negative acceleration so when giving the value of deceleration it must be positive as it already means negative acceleration. d) Total distance = area under the graph. The graph is a trapezium so its area = 25 (3 + 6) = 112.5. So the distance is 112.5 m.
IMPORTANT:
When you are asked to describe motion you should refer to acceleration (or both acceleration of the car and its speed). When you are asked to describe the speed of the car just refer to its speed not to acceleration. If you refer to acceleration your answer will be wrong. When you are asked to describe the acceleration of the car then refer to its acceleration only. Do not refer to its speed at all. For this reason I have described both acceleration and speed on the v / t graphs above.
By Shafaq Hafeez shafaq@physics.com.pk