14 Chapter 1 • Dynamics
Refer to equation 1.1 {g = 9,8 m/s2} and the following example of a free-falling body. Example 1.12
Consider a body from the moment it is released: Aſter 1 second, Aſter 2 seconds, Aſter 3 seconds, Aſter 4 seconds,
v1 = 9,8 m/s
v2 = 9,8 + 9,8 = 19,6 m/s v3 = 19,6 + 9,8 = 29,4 m/s
v4 = 29,4 + 9,8 = 39,2 m/s and so on.
Te velocity/time graph for a body with uniform acceleration (such as a free-falling body) is a straight line. For a body initially at rest, t = 0 and v = 0. After t = 1 s, g = 9,8 m/s2. Tis can be plotted on a graph (fig. 1.6) with scales of 2 cm = 1 s and 2 cm = 10 m/s, as follows: y
40 39,2 (d) 30 20 10 3y 19,6 (c) 9,8 (a) 0 1 Fig. 1.6 Velocity/time graph
acceleration of the body: g = 3
3 g. 1.6 Velocity/time graph Te slope (gradient) of a straight line graph is 3 3
y x . In this case it would represent the
v t m/s2 ................................................................................................................ (1.6)
This verifies the definition of acceleration. Refer to definition 1.10, Acceleration.
Example 1.12 can also be illustrated by this graph. Fromthe graph: After 2 seconds (Oa), v = 19,6 m/s (Oc) After 4 seconds (Ob), v = 39,2 m/s (Od)
Time (t) in s 2 3 (b) 4 x
1444442444443 3x
Velocity (v) in m/s
1444442444443
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