The document provides information about displacement-time graphs and velocity-time graphs including: 1) Examples of different types of displacement-time and velocity-time graphs showing various motions like constant velocity, acceleration, deceleration, and vertical motion. 2) Definitions of velocity, acceleration, and the relationships between the gradients of the graphs and these quantities. 3) Kinematics equations for uniformly accelerated motion derived from the area under the velocity-time graph.

1. flipperworks.com
Examples of displacement-time graphs
K v
2. highest point s
i (c) s is max, v = 0
n
s v
e
m 3. body falling.
1. As body
a goes up, s ↑ s ↓ but v ↑ in
t but v ↓. magnitude
i
c t
s motion of a body thrown vertically up
and then returning to the point of projection

2. flipperworks.com
Velocity and Acceleration
Definition Graphically
Velocity Rate of ds Gradient of s-t
change of v= graph
displacement dt
Acceleration Rate of Gradient of v-t
dv
change of
velocity
a= graph
dt

3. flipperworks.com
K [b] Velocity-time graph
i
n  shows the velocity of a body at any
e instant of time.
t
i  the gradient of the graph is the
c instantaneous acceleration of the body.
s
 the area under the graph is the
displacement of the body.

4. flipperworks.com
v
B C
A
Area = A1
O E H t
Area = A2
D
F G
v (+ve when pointing to the right)
O

5. flipperworks.com
v
B C
A
Area = A1
O E H t
Area = A2
D
F G
O to A
v increases from zero at constant
rate
⇒ body is moving from rest with
uniform acceleration

6. flipperworks.com
v
B C
A
Area = A1
O E H t
Area = A2
D
F G
A to B
v increasing but gradient decreasing
⇒ body continues to move faster but
with decreasing acceleration.

7. flipperworks.com
v
B C
A
Area = A1
O E H t
Area = A2
D
F G
B to C
v remains constant
⇒ zero acceleration

8. flipperworks.com
v
K
B C
i A
n Area = A1
e E H
O t
m D Area = A2
c F G
s
C to D
v decreases at a constant rate but still
positive
⇒ acceleration is constant but negative
i.e. constant deceleration.

9. flipperworks.com
v
B C
A
Area = A1
O E H t
Area = A2
D
F G
D to E
v = 0 ⇒ body is stationary

10. flipperworks.com
v
Q: At time F, is
B C displacement of car
A negative?
Area = A1
O E H t
Area = A2
D
F G
E to F
v negative but magnitude is increasing at
constant rate ⇒ body is moving in opposite
direction and speeds up ⇒ uniform negative
acceleration.

11. flipperworks.com
v
Q: At time F, is
B C displacement of car
A negative?
Area = A1
O E H t
Area = A2
D
F G
v (+ve when pointing to the right)
O A B C D
HG F E

12. flipperworks.com
v
K
B C
i A
n Area = A1
e E H
O t
a D Area = A2
t F G
i
c F to G
s v remains constant

13. flipperworks.com
v
B C
A
Area = A1
O E H t
Area = A2
D
F G
G to H
v is negative but acceleration is
positive ⇒ constant deceleration.
Body slows down and comes to rest
at H

14. flipperworks.com
v
B C
A
Area = A1
O E H t
Area = A2
D
F G
Total distance moved = A1 + A2
Net displacement = A1 - A2

15. flipperworks.com
Examples of velocity-time graphs
(a) v (b) v
2
3
1
t t
line 1: uniform velocity increasing acceleration
line 2: uniform acceleration
line 3: uniform deceleration

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Q: Can it be ball thrown
v upwards, hitting ceiling
(c)
and bouncing back?
0 t
Ball, released from rest at a certain height,
hitting the floor and bouncing back

17. flipperworks.com
(c) +
v
v2
g
0 t v2
-v1 v1
free fall ⇒ acceleration = gradient = -g

18. flipperworks.com
Example 2:
v/ s-1
m Velocity after 10 s
= 0.80 x 10
= 8.0 m s-1
8.0
total distance travelled
= area under graph
= ½(20 +30) 8.0
t/
s = 200 m
0 10 30

19. flipperworks.com
Example 3:
The graph shows the variation with time of the
velocity of a trolley, initially projected up an
inclined runway.
Velocity/m s-1
0.8
0.6
0.4
0.2
θ
0 Time/
s
-0.2
-0.4
-0.6
-0.8

20. flipperworks.com
(a) maximum distance
= area under v-t graph between t = 0 & t = 2.5 s
= ½ × 0.80 × 2.5 = 1.0 m
Velocity/m s-1
0.8
0.6
0.4 Trolley reaches
0.2 max. distance
0 Time/ velocity = 0
when
s
-0.2
-0.4
-0.6
-0.8

21. flipperworks.com
(b) a = gradient of v-t graph
0.00 - 0.80
= = - 0.32 m s-2
2.5 - 0.0
∴ deceleration = 0.32 m s-2
Velocity/m s-1
0.8
0.6
0.4
0.2
0 Time/
s
-0.2
-0.4
-0.6
-0.8

22. flipperworks.com
(c) displacement = total area under graph
= 1.0 + (-1.0)
=0m
Velocity/m s-1
0.8
0.6
0.4
0.2 +1.0
0 Time/
s
-0.2
-0.4 -1.0
-0.6
-0.8

23. flipperworks.com
(d) Trolley travels 1.0 m up the runway with
uniform deceleration, stops momentarily at
t = 2.5 s and then accelerates uniformly
down the runway.
Velocity/m s-1
0.8
0.6
0.4
0.2
0 Time/
s
-0.2
-0.4
-0.6
-0.8

24. flipperworks.com
Displacement
Speed
Velocity
Acceleration
Average speed
Average velocity
WORDS &
TERMS
KINEMATICS
EQUATIONS
GRAPHS

25. flipperworks.com
Equations representing uniformly
accelerated motion in a straight line
Suppose that a body is moving with constant
acceleration a and that in a time interval t, its
velocity increases from u to v and its
displacement increases from 0 to s .
v
u
0 t
Since a = d v / d t ⇒ a = v - u
t
Hence v = u + at -------------- (1)

26. flipperworks.com
Since velocity increases steadily,
u+v
average velocity, < v > =
2
Recall: displacement, s = average velocity × time
Thus, s = ½ (u + v) t ------------- (2)
Substituting (1) into (2),
s = ut + ½ at2 -------------- (3)

27. flipperworks.com
v -u
From (1), t =
a
Substituting this into (2),
u+v v -u
s = ×
2 a
Therefore v 2 = u 2 + 2 a s -------- (4)

28. flipperworks.com
Kinematics equations for uniformly
accelerated motion in a straight line :
v = u +at
s = ½ (u + v) t recall & derive
s = u t + ½ a t2
v2 = u2 + 2 a s
Since u, v , a and s are vector quantities,
their directions must be taken into
account when solving problems.

29. flipperworks.com
Sign Conventions
Eg. A ball is released from a certain height.
Starting
+ position
s is -ve
a is -ve
v is -ve