Environmental Engineering Reference
In-Depth Information
1.7 The height of the building is 216 m. The flight time is 6.63 s.
1.8 The velocity is obtained by differentiation, which is straightforward since the
basis vectors are constant in time:
d
s
d
t
0
.
01
t
k
km s
−
1
.
v
=
=
0
.
3
i
+
0
.
5
j
−
√
0
.
3
2
0
.
58 km s
−
1
.
=
|
v
| =
+
0
.
5
2
=
With
t
0,
The
velocity
at
this
√
0
.
3
2
|
|
=
0
.
5
2
0
.
3
2
time
lies
in
the
xy
plane.
At
t
=
30 s,
v
+
+
0
.
66 km s
−
1
. To work out the angle to the vertical use
=
v
z
|
cos
θ
=
v
|
117
◦
.
1.9 Running with the travelator, velocities add so
v
which will give
θ
=
11 ms
−
1
,against
v
=
=
9ms
−
1
so the total time for both legs is
50
1
1
9
t
=
11
+
=
10
.
1s
.
Since time is inversly proportional to speed the contribution of the motion of
the travelator does not cancel over the two legs in the way that the athelete
expected it would.
1.10 To cross the river she must have a net velocity
v
that lies on the line joining
the two stations. She must therefore sail with a velocity
v
relative to the
river, such that
v
=
+
v
u
where
u
is the velocity of the river. Since the three vectors form a right-angled
triangle the time taken to cross is
d
√
v
2
t
=
u
2
.
−
There is no real solution if
u>v
making the crossing impossible in this case.
PROBLEMS 2
(a) The equation for the position vector
r
for a general point on the line
through
r
1
and
r
2
can be written
2.1
sfasfd
r
−
r
1
=
λ(
r
2
−
r
1
)
where
λ
is real. When 0
1
r
lies between
r
1
and
r
2
.Fromthe
definition of the centre-of-mass of two particles you can obtain
≤
λ
≤
m
1
m
1
+
R
−
r
1
=
(
r
2
−
r
1
)
m
2
