16
1 Kinematics and Statics
v =
dx
dt
= 6t − 12
v = 0 gives t = 2 s
(3)
Using (3) in (2) gives displacement x = 0
1.7 s = ut +
1
2
at
2
(1)
∴ h = u × 2 −
1
2
g × 2
2
(2)
h = u × 10 −
1
2
g × 10
2
(3)
Solving (2) and (3) h = 10g = 10 × 9.8 = 98 m.
1.8 Take the origin at the position of A at t = 0. Let the car A overtake B in time t
after travelling a distance s. In the same time t, B travels a distance (s − 30) m:
s = ut +
1
2
at
2
(1)
s = 13t +
1
2
× 0.6 t
2
(Car A)
(2)
s − 30 = 20t −
1
2
× 0.46 t
2
(Car B)
(3)
Eliminating s between (2) and (3), we find t = 0.9 s.
1.9 Let BD = x. Time t 1 for crossing the field along AD is
t 1 =
AD
v 1
=
x 2 + (600) 2
1.0
(1)
Time t 2 for walking on the road, a distance DC, is
t 2 =
DC
v 2
=
800 − x
2.0
(2)
Total time t = t 1 + t 2 =
x 2 + (600) 2 +
800 − x
2
(3)
Minimum time is obtained by setting dt/dx = 0. This gives us x = 346.4 m.
Thus the boy must head toward D on the round, which is 800–346.4 or 453.6 m
away from the destination on the road.
The total time t is obtained by using x = 346.4 in (3). We find t = 920 s.
1.10 Time taken for the first drop to reach the floor is
t 1 =
2 h
g
=
2 × 2.45
9.8
=
1
√
2
s