Everything Science Grade 10

21.6 CHAPTER 21. MOTION IN ONE DIMENSION

0

20

40

60

80

100

0 20 40 60 80 100

position

x

(m)

timet(s)

b

∆t

b ∆x

b

− 1

− 2

20 40 60 80 100

velocity

v

(m

−·s

1 )

timet(s)

b b b

0

1

2

0 20 40 60 80 100

acceleration

a

(m

−·s

2 )

timet(s)

b b b

Graphs for motion with a constant negative velocity. The area of the shaded portion in the
vvs.tgraph corresponds to the object’s displacement.

We see that the~vvs.tgraph is a horizontal line. If the velocity vs. time graph is a horizontal
line, it means that the velocity isconstant(not changing). Motion at a constant velocity is
known asuniform motion. We can use the~xvs.tto calculate the velocity by finding the
gradient of the line.

v=∆∆~xt

=~xtf−~xi

f−ti

=^0100 m−s^100 − 0 sm

=− 1 m·s−^1

Vivian has a velocity of− 1 m·s−^1 , or 1 m·s−^1 towards her house. You will notice that the
~vvs.tgraph is a horizontal line corresponding to a velocity of− 1 m·s−^1. The horizontal
line means that the velocity stays the same (remains constant) during the motion. This is
uniform velocity.

We can use the~vvs.tto calculate the acceleration by finding the gradient of the line.

a=∆∆~vt

=~vtf−~vi

f−ti

=^1 m·s

− (^1) − 1 m·s− 1
100 s− 0 s
= 0m·s−^2
Vivian has an acceleration of 0 m·s−^2. You will notice that the graph of~avs.tis a horizontal
line corresponding to an acceleration value of 0 m·s−^2. There is no acceleration during
the motion because his velocity does not change.
We can use the~vvs.tgraph to calculate the displacement by finding the area under the
graph.
410 Physics: Mechanics