1.3 Solutions
23
It follows that t 2 > t 1 , that is, time of descent is greater than the time of ascent.
Further, from (4) and (6)
v
u
=
g −
F
m
g +
F
m
(9)
It follows that v < u, that is, the final speed is smaller than the initial speed.
1.22 Taking the downward direction as positive, the equation of motion will be
dv
dt
= g − kv
(1)
where k is a constant. Integrating
dv
g − kv
=
dt
∴ −
1
k
ln
g − kv
c
= t
where c is a constant:
g − kv = ce
−kt
(2)
This gives the velocity at any instant.
As t increases e −kt decreases and if t increases indefinitely g − kv = 0, i.e.
v =
g
k
(3)
This limiting velocity is called the terminal velocity. We can obtain an expression for the distance x traversed in time t. First, we identify the constant c
in (2). Since it is assumed that v = 0 at t = 0, it follows that c = g.
Writing v =
dx
dt
in (2) and putting c = g, and integrating
g − k
dx
dt
= ge
−kt
gdt − k
dx = g
e
−kt dt + D
gt − kx = −
g
k
e
−kt
+ D
At x = 0, t = 0; therefore, D =
g
k
23
It follows that t 2 > t 1 , that is, time of descent is greater than the time of ascent.
Further, from (4) and (6)
v
u
=
g −
F
m
g +
F
m
(9)
It follows that v < u, that is, the final speed is smaller than the initial speed.
1.22 Taking the downward direction as positive, the equation of motion will be
dv
dt
= g − kv
(1)
where k is a constant. Integrating
dv
g − kv
=
dt
∴ −
1
k
ln
g − kv
c
= t
where c is a constant:
g − kv = ce
−kt
(2)
This gives the velocity at any instant.
As t increases e −kt decreases and if t increases indefinitely g − kv = 0, i.e.
v =
g
k
(3)
This limiting velocity is called the terminal velocity. We can obtain an expression for the distance x traversed in time t. First, we identify the constant c
in (2). Since it is assumed that v = 0 at t = 0, it follows that c = g.
Writing v =
dx
dt
in (2) and putting c = g, and integrating
g − k
dx
dt
= ge
−kt
gdt − k
dx = g
e
−kt dt + D
gt − kx = −
g
k
e
−kt
+ D
At x = 0, t = 0; therefore, D =
g
k