Information about, Motion and Distance time graph

1. MOTION
GRAPHICAL REPRESENTATION OF
MOTION

2. INTRODUCTION:
 Graphs provide us a convenient way to present basic information about
various events.
 For example:
Run rate shown in cricket matches. Used to solve equations in two
variables.

3. USAGE OF GRAPHS:
 We can use line graphs to describe motion. We can show the dependence of
one physical quantity on other through line graphs.
 For example:
Distance-time graphs Velocity-time graphs
 Uniform velocity and uniform speed are equal when the magnitude of distance
and displacement is equal.

4. SIMPLE FLOW CHART TO
UNDERSTAND THE TOPIC:
GRAPHICAL REPRESENTATION OF MOTION
DISTANCE-TIME GRAPHS VELOCITY-TIME GRAPHS
CALCULATE SPEED USING
DISTANCE AND TIME
CALCULATE DISTANCE /d /a USING
VELOCITY AND TIME

5. DISTANCE-TIME GRAPHS:
 In a distance-time graph the distance is represented along the “y-axis” and
time is represented along the “x-axis”.
 Using a distance-time graph the change in the position of an object with
respect to time can be determined.
 A distance-time graph helps us find the speed of the object moving.
 When the object is under going uniform motion(velocity) the graph obtained
will always be a straight line graph, as the distance and time are directly
proportional to each other
 When the object is under going non-uniform motion(velocity) the graph
obtained will not be a straight line graph.
 When we are finding the speed we have to use the concept of slope.

6. HOW TO DETERMINE THE
SPEED FROM THIS GRAPH:
 To determine speed using the graph we have to:
1. Take two consecutive points and draw perpendiculars to the x & y axis as
shown in the figure.
2. Name the two points ‘A’ & ‘B’.
3. Name the point on the y-axis as S1 and the point on the x-axis asT1 (from
point A).
4. Name the point on the y-axis as S2 and the point on the x-axis as T2 (from
point B).
5. Speed =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
 Speed =
𝑦−𝑎𝑥𝑖𝑠
𝑥−𝑎𝑥𝑖𝑠
=slope  Slope =
𝑦2
−𝑦1
𝑥2
−𝑥1
 Slope =
𝑠2
−𝑠1
𝑡2
−𝑡1

7. HOW TO SOLVE:
1. Here s1=30km, s2= 40km; t1=60mins,t2=80mins.
2. So, according to the formula- Slope =
𝑠2
−𝑠1
𝑡2
−𝑡1
Slope =
𝑦2
−𝑦1
𝑥2
−𝑥1
 Slope =
𝑠2
−𝑠1
𝑡2
−𝑡1
=
40−30
80−60
=
10
20
=
1
2
=0.5x60
Therefore, according to this diagram the vehicle is moving with a speed of 30
km/hr.

8. NON-UNIFORM SPEED(VELOCITY)
 The nature of this graph shows nonlinear variation of the distance travelled
by the car with time. Thus, the graph shown in represents motion with non-
uniform speed.

9. VELOCITY-TIME GRAPH:
 In a velocity-time graph the velocity is represented along the “y-axis” and the
time is represented along the “x-axis”.
 Whenever we fix a velocity on the y-axis its always a straight line parallel line
to the x-axis.
 The velocity is always independent of time i.e. in whichever time we try to find
the velocity the velocity remains the same.
 When the object has an uniform or constant velocity the graph obtained will
always be a straight line graph.
 Whenever the object does not have an uniform velocity the graph obtained
will be a curved line graph.
 When we have to find the distance we cannot use the concept of slope as
slope uses the concept of division but [D=SxT].
 When we are finding the distance we have to find the area under the graph.

10. HOW TO DETERMINE DISTANCE
FROM THIS GRAPH:
 To determine distance from this graph we have to:
1. (If the velocity is a constant) draw a straight line parallel to the x-axis.
2. Then take two consecutive points and name them ‘A’ & ‘B’.
3. Then draw two perpendicular lines from point ‘A’ & ‘B’ and name them ‘C’ & ‘D’.
4. Mark the two points ‘C’ & ‘D’ as T1 & T2 respectively.
5. We have to find the area of the rectangle ABDC=lxb.

11. HOW TO SOLVE:
1. LxB = ACxCD
2. Here the distance=ACxCD.
3. Where AC=40m/s,CD=6secs.
4. Therefore, ACxCD=40x6=240m.
5. Which implies that according to this diagram the moving vehicle has covered a
distance of 240m.

12. HOW TO DETERMINE DISTANCE
USING THIS GRAPH:
 Scale y-axis=10m/s for 1 unit; x-axis=2secs for 1 unit.
 In this figure we can calculate the distance in two ways
1. Area of trapezium ABDE or
2. Area of triangle ABC+Area of rectangle ACDE.

13. HOW TO SOLVE:
 Using area of trapezium:
1. Area of trapezium=
(𝑆𝑢𝑚 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑠𝑖𝑑𝑒𝑠) 𝑋 ℎ𝑒𝑖𝑔ℎ𝑡
2
2. So, in the above figure: AEllBD and AC=height.
3. Therefore,
(𝐴𝐸+𝐵𝐷)𝑋𝐴𝐶
2
=
(25+55)𝑋8
2
=
640
2
= 320 km
4. Thus, the vehicle in this case has travelled a distance of 320 km.

14. HOW TO SOLVE:
 Using area of triangle+area of rectangle=1/2xbxh+lxb
 Area of triangle-
1.
1
2
xbxh =
1
2
xACxBC
2.
1
2
x30x8 pn: Change in colour represents cancellation.
3. 30x4=120 km
 Area of rectangle-
1. lxb=25x8
2. 25x8=200 km
 Therefore the total distance=120+200=320 km.

15. HOW TO DETERMINE DISTANCE
USING THIS GRAPH:
 In this case to find the distance we have to find the area under the graph
which is the area of the triangle.
 Area of triangle=
 Here we can take a point anywhere on the graph and we have to find the are
between the origin and the point taken.
 We have taken the point as ’A’ on the perpendiculars of 8 and 2. The origin is
‘O’ on zero and the other point on two as ‘C’.
 So we have to find the area of the triangle AOC.
 The difference between the previous graph and this graph is just the marking
of points, in the 1st one it’s any 2 points on the graph whereas here in the 2nd
one it only one point on the graph and your other point is nothing but the
origin.
1
2
xbxh

16. HOW TO SOLVE:
1. Area of triangle=
1
2
xbxh
2.
1
2
xbxh=
1
2
x2x8
3. Therefore, the area of the triangle AOC=8m2
4. Thus, the vehicle has travelled a distance of 8m.

17. VELOCITY-TIME GRAPHS OF AN OBJECT
IN NON-UNIFORMLY ACCELERATED
MOTION:
 (a) shows a velocity-time graph that represents the motion of an object whose
velocity is decreasing with time while (b) shows the velocity-time graph
representing the non-uniform variation of velocity of the object with time.

18. TYPES OF DISTANCE-TIME GRAPHS:
 If the graph obtained is a
straight line, then the object is
said to moving in increasing
uniform motion
 If the graph obtained is a
straight line parallel to the
time axis, then object is in
rest.
If the graph obtained is a
straight line as shown, then
the object is said to moving in
uniform motion and is moving
towards the initial position

19.  If the graph obtained is a curved
line as shown, then the object is
said to be in increasing non uniform
motion
 If the graph obtained is a curved
line as shown, then the object is
said to be in decreasing non
uniform motion

20. TYPES OF VELOCITY TIME GRAPHS:
 If the graph obtained is a
straight line parallel to the
time axis, then the object is
said to be with uniform
velocity. Therefore the
object is having zero
acceleration
 If the graph obtained is a
straight line as shown, then the
body is said to have uniform
acceleration.

21. RETARDATION: NEGATIVE
ACCELERATION:
 If the graph is a straight
line as shown, then the
object is said to be moving
with uniform retardation
 If the graph is a curved
line as shown, then the
object is said to be
moving with non uniform
acceleration
.
If the graph is a curved
line as shown, then the
object is said to be
moving with non uniform
retardation

22. VELOCITY TIME GRAPH
 In this graph, the object
accelerates in AB, moves with
constant velocity and zero
acceleration till BC and retardation
occurs till CD
DISTANCE TIME GRAPH
 In this graph, the object covers
distance in uniform speed in AB,
stays in rest in BC and then comes
back to the initial position.

23. CLUES FOR V-T GRAPHS:

24. SHORT SUMMARY OF THE TOPIC:
 When an object covers equal distance in equal intervals of time – Uniform
velocity.
 When an object covers unequal distances in equal intervals of time – Non-
uniform velocity.
 From the distance-time graph the speed is given by the slope of the line.
 From the velocity-time graph the distance is found by the area under the
graph or the area enclosed by the figure.

25. NUMERICALS:
1. Find the speed of the vehicle between
A and B.
Sol: Speed=Slope=S2-S1/ T2-T1
=4-3/6-4
=1/2
=0.5m/s
A
B

26. NUMERICALS:
1. Find the area of the shaded region.
Sol: Area of trapezium=(Sum of ll sides) x height/2
=(20+50)x3/2
=(70)x3/2
=210/2
=105m

27. NUMERICALS:
1. Find the distance covered by finding the shaded region.
Sol: Area of rectangle=lxb
=l=30,b=6
=lxb=30x6
=180m

28. NUMERICALS:
1. Find the area of the shaded region to find the distance.
Sol: Area of the shaded region=Area of a triangle
=1/2xbxh
=1/2x4x40
=1x2x40
=80m

29. NUMERICALS:
1. Find the speed of the car between the perpendicular of
(10,1) and (50,5).
Sol: Slope=Speed=S2-S1/T2-T1
=50-10/5-1
=40/4
=10 m/s

30. MCQ’S:
1. Distance-time graph of two objects A and B are shown below. Which
statement is true for the speed of object A and B?
a) Speed of object A is greater than object B
b) Speed of object A is lesser than object B
c) Both have same speed
d) Speed of Object A is double the speed of object B.

31. MCQ’S
1. In a distance-time graph, if the line is horizontal, then the object is:
a) Accelerating
b) Speeding up
c) At rest
d) Slowing down

32. MCQ’S
1. On the distance-time graph, the Y-axis should be labelled as:
a) Distance
b) Displacement
c) Speed
d) Time

33. MCQ’S
1. The velocity-time graph of an object moving in a fixed direction is shown in
the figure below. What do you conclude about the object?
a) Moves with a constant speed.
b) Moves with varying speeds.
c) Moves with a non zero acceleration.
d) Is at rest.

34. MCQ’S
1. The slope of the distance-time graph is:
a) Distance
b) Acceleration
c) Speed
d) Displacement

35. MCQ’S
1. The area enclosed by velocity-time graph and the time axis will be equal to the
magnitude of:
a) Velocity
b) Speed
c) Acceleration
d) Distance

36. MCQ’S
1. For a constant acceleration, the nature of velocity-time graph is:
a) Graph-B
b) Graph-D
c) Graph-C
d) Graph-A

37. MCQ’S
1. Area under a v – t graph represents a physical quantity which has the unit:
a) m2
b) m/s
c) m/s2
d) m

38. MCQ’S
1. The acceleration of an object moving in a straight line can be determined
from:
a) The area between the distance-time graph and time axis.
b) The area between the velocity-time graph and time axis
c) The slope of the velocity-time graph.
d) The slope of the distance-time graph.

39. MCQ’S
1. This is a velocity-time graph when the body is under:
a) Non-uniform retardation.
b) Uniform acceleration.
c) Uniform retardation.
d) Non-uniform acceleration.

40. DONE BY: DIVYA & SHRIMAYI