1.3 Solutions
19
Fig. 1.16
W 2 = W − 2v
By problem angle CAD = θ = 45 ◦ . The triangle ACD is therefore an isosceles right angle triangle:
AD =
√
2CD = 4
√
2 km/h
Therefore the actual speed of the wind is 4
√
2 km/h from southeast direction.
1.17 Choose the floor of the elevator as the reference frame. The observer is inside
the elevator. Take the downward direction as positive.
Acceleration of the bolt relative to the elevator is
a
= g − (−a) = g + a
h =
1
2
a
t
2
=
1
2
(g + a)t
2
t =
2h
g + a
1.18 In 2 s after the truck driver applies the brakes, the distance of separation
between the truck and the car becomes
d rel = d −
1
2
at
2
= 10 −
1
2
× 2 × 2
2
= 6 m
The velocity of the truck 2 becomes 20 − 2 × 2 = 16 m/s.
Thus, at this moment the relative velocity between the car and the truck will be
u rel = 20 − 16 = 4 m/s
Let the car decelerate at a constant rate of a 2 . Then the relative deceleration
will be
a rel = a 2 − a 1
19
Fig. 1.16
W 2 = W − 2v
By problem angle CAD = θ = 45 ◦ . The triangle ACD is therefore an isosceles right angle triangle:
AD =
√
2CD = 4
√
2 km/h
Therefore the actual speed of the wind is 4
√
2 km/h from southeast direction.
1.17 Choose the floor of the elevator as the reference frame. The observer is inside
the elevator. Take the downward direction as positive.
Acceleration of the bolt relative to the elevator is
a
= g − (−a) = g + a
h =
1
2
a
t
2
=
1
2
(g + a)t
2
t =
2h
g + a
1.18 In 2 s after the truck driver applies the brakes, the distance of separation
between the truck and the car becomes
d rel = d −
1
2
at
2
= 10 −
1
2
× 2 × 2
2
= 6 m
The velocity of the truck 2 becomes 20 − 2 × 2 = 16 m/s.
Thus, at this moment the relative velocity between the car and the truck will be
u rel = 20 − 16 = 4 m/s
Let the car decelerate at a constant rate of a 2 . Then the relative deceleration
will be
a rel = a 2 − a 1