From 100 to 120 ms, region
C
has the shape of a rectangle with area
2
C
area
(0.0200 s) (50.0 m/s
) = 1.00 m/s.
=
From 110 to 160 ms, region
D
has the shape of a trapezoid with area
2
D
1
area
(0.0400 s) (50.0
20.0) m/s
1.40 m/s.
2
=
+
=
Substituting these values into Eq. 226, with
v
0
=0 then gives
1
0
0
1 50 m/s + 1.00 m/s + 1.40 m/s = 3.90 m/s,
v
.
−
=
+
or
1
3 90 m/s.
v
.
=
65. The key idea here is that the speed of the head (and the torso as well) at any given
time can be calculated by finding the area on the graph of the head’s acceleration versus
time, as shown in Eq. 226:
1
0
0
1
area between the acceleration curve
and the time axis, from
o
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Stoler
 Physics, Acceleration, Harshad number, m/s

Click to edit the document details