ABSTRACT
Problem 1. A car starts from rest and its speed is measured every second for 6 s:
Time t(s) 0 1 2 3 4 5 6 Speed v
(m/s) 0 2.5 5.5 8.75 12.5 17.5 24.0 Determine the distance travelled in 6 seconds (i.e. the area under the v/t graph), by (a) the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson’s rule
A graph is shown in
(a) Trapezodial rule (see para. (b) above). The time base is divided into 6 strips each of width 1 s, and the length of the ordinates measured. Thus
area = (1) [(
0 + 24.0 2
) + 2.5 + 5.5 + 8.75
+12.5 + 17.5 ]
= 58.75 m
(b) Mid-ordinate rule (see para. (c) above). The time base is divided into 6 strips each of width 1 second. Mid-ordinates are erected as shown in Fig. 21.3 by the broken lines. The length of each mid-ordinate is measured. Thus
area = (1)[1.25 + 4.0+ 7.0 + 10.75 + 15.0 + 20.25] = 58.25 m
(c) Simpson’s rule (see para. (d) above). The time base is divided into 6 strips each of width 1 s, and the length of the ordinates measured. Thus
area = 1 3 (1)[(0 + 24.0)+ 4(2.5 + 8.75 + 17.5)+ 2(5.5 + 12.5)] = 58.33 m
Problem A the depth are made at equal intervals of 3 m across the river and are as shown below.
