1.3 Solutions
27
y = 4t +
t 2
2
+ D
y = 0, t = u; D = u
y = u + 4t +
t 2
2
(ii)
v = (6 + 2t) ˆ
i + (4 + t) ˆ
j
(iii)
a =
dv
dt
= 2 ˆ
i + ˆ
j
(iv) a =
2 2 + 1 2 =
√
5
tan θ =
1
2
; θ = 26.565
◦
Acceleration is directed at an angle of 26 ◦ 34 with the x-axis.
1.27 Take upward direction as positive, Fig. 1.18. At time t the velocities of the
objects will be
v 1 = u 1 ˆ
i − gt ˆ
j
(1)
v 2 = −u 2 ˆ
i − gt ˆ
j
(2)
If v 1 and v 2 are to be perpendicular to each other, then v 1 · v 2 = 0, that is
u 1 ˆ
i − gt ˆ
j
·
−u 2 ˆ
i − gt ˆ
j
= 0
∴
−u 1 u 2 + g
2 t
2
= 0
or
t =
1
g
√
u 1 u 2
(3)
The position vectors are r 1 = u 1 t ˆ
i −
1
2 gt 2 ˆ
j, r 2 = −u 2 t ˆ
i −
1
2 gt 2 ˆ
j.
The distance of separation of the objects will be
r 12 = | r 1 − r 2 | = (u 1 + u 2 )t
Fig. 1.18