20
1 Kinematics and Statics
If the rear-end collision is to be avoided the car and the truck must have the
same final velocity that is
v rel = 0
Now v
2
rel = u
2
rel − 2 a rel d rel
a rel =
v 2
rel
2 d rel
=
4 2
2 × 6
=
4
3
m/s
2
∴ a 2 = a 1 + a rel = 2 +
4
3
= 3.33 m/s
2
1.19 v BA = v B − v A
From Fig. 1.17a
v BA =
v 2
B + v 2
A − 2v B v A cos 60 ◦
=
20 2 + 30 2 − 2 × 20 × 30 × 0.5 = 10
√
7 km/h
The direction of v BA can be found from the law of sines for ABC,
Fig. 1.17a:
(i)
AC
sin θ
=
BC
sin 60
or sin θ =
AC
BC
sin 60 =
v B
v BA
sin 60
◦
=
20 × 0.866
10
√
7
= 0.6546
θ = 40.9
◦
Fig. 1.17a
1 Kinematics and Statics
If the rear-end collision is to be avoided the car and the truck must have the
same final velocity that is
v rel = 0
Now v
2
rel = u
2
rel − 2 a rel d rel
a rel =
v 2
rel
2 d rel
=
4 2
2 × 6
=
4
3
m/s
2
∴ a 2 = a 1 + a rel = 2 +
4
3
= 3.33 m/s
2
1.19 v BA = v B − v A
From Fig. 1.17a
v BA =
v 2
B + v 2
A − 2v B v A cos 60 ◦
=
20 2 + 30 2 − 2 × 20 × 30 × 0.5 = 10
√
7 km/h
The direction of v BA can be found from the law of sines for ABC,
Fig. 1.17a:
(i)
AC
sin θ
=
BC
sin 60
or sin θ =
AC
BC
sin 60 =
v B
v BA
sin 60
◦
=
20 × 0.866
10
√
7
= 0.6546
θ = 40.9
◦
Fig. 1.17a