Contoh Soal Aplikasi 62292h
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CONTOH SOAL APLIKASI 1. A particle falls in a vertical line under gravity (supposed constant) and the force of air resistance to its motion to its velocity. Show that its velocity cannot exceed a particular limit. Solution : let V be the velocity when particle has fallen a distance S in time t from rest. If the resistance is kV, then the equation of motion is :
(
Integrating
)
Initial condition = V = 0 at = t = 0 C= (
becomes
)
t= (
V=
)
Limit velocity or maximum velocity V = 2. The acceleration an velocity of a body falling in the air approximately satisfy the equation: Acceleration = g – kV2 where v is the velocity of the body at any time t and k, g are constant. Find the distance traversed as a function of the time t, if the body fall from rest. Show that the value of will never exceed √ Solution : acceleration = g – kV2 Or Or
√
[
√
√
√
√
]dv = dt
On integrating we get √ √
(√
√
√
)
√ √
√
√
√
√
(√
√
)
When t = 0 v = 0 √
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
By component and dividend √
√
√
√
√
√
√
√
√
√
Or v = √
Whatever the value of t may be tanh √ Hence the value of v will never exceed √ √
√
Integrating again x=
∫
x= when t = 0, x x=
√ √ D=0 √
ans
3. A moving body is opposed by a force per unit mass of value ex and resistance per unit mass of value bv2 where x and v are displacement and velocity of the particle at the instant. Find the velocity of the particle in of x, if it stars from the rest.
Solution : by Newton second laws of motion, the equation of motion of body is
Putting v2 = z,2v =
I.F =
∫
z.
=∫ ]
∫
= -2c[ =z=v=-
initially
where
0= z=
ans
4. The rate a which a body cools is proportional to the difference between the temperature of the body and that of surrounding air. If a body air at 25 0 will cool from 1000 to 750 in one minute, find its temperature at the end off three minutes Solution : let temperature of the body be T 0 C (
)
= kdt
log (T – 25 ) = kt + log A or log
= kt
T – 25 = When t = 0 then T = 100 from (1) A = 75 When t = 1 then T = 25 and A = 75 from (1) = ek
(1) becomes T = 25 + 75 ekt When t = 3 then T = 25 + 75 ekt = 25 + 75 x 8/27 = 47,22
(
Integrating
)
Initial condition = V = 0 at = t = 0 C= (
becomes
)
t= (
V=
)
Limit velocity or maximum velocity V = 2. The acceleration an velocity of a body falling in the air approximately satisfy the equation: Acceleration = g – kV2 where v is the velocity of the body at any time t and k, g are constant. Find the distance traversed as a function of the time t, if the body fall from rest. Show that the value of will never exceed √ Solution : acceleration = g – kV2 Or Or
√
[
√
√
√
√
]dv = dt
On integrating we get √ √
(√
√
√
)
√ √
√
√
√
√
(√
√
)
When t = 0 v = 0 √
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
By component and dividend √
√
√
√
√
√
√
√
√
√
Or v = √
Whatever the value of t may be tanh √ Hence the value of v will never exceed √ √
√
Integrating again x=
∫
x= when t = 0, x x=
√ √ D=0 √
ans
3. A moving body is opposed by a force per unit mass of value ex and resistance per unit mass of value bv2 where x and v are displacement and velocity of the particle at the instant. Find the velocity of the particle in of x, if it stars from the rest.
Solution : by Newton second laws of motion, the equation of motion of body is
Putting v2 = z,2v =
I.F =
∫
z.
=∫ ]
∫
= -2c[ =z=v=-
initially
where
0= z=
ans
4. The rate a which a body cools is proportional to the difference between the temperature of the body and that of surrounding air. If a body air at 25 0 will cool from 1000 to 750 in one minute, find its temperature at the end off three minutes Solution : let temperature of the body be T 0 C (
)
= kdt
log (T – 25 ) = kt + log A or log
= kt
T – 25 = When t = 0 then T = 100 from (1) A = 75 When t = 1 then T = 25 and A = 75 from (1) = ek
(1) becomes T = 25 + 75 ekt When t = 3 then T = 25 + 75 ekt = 25 + 75 x 8/27 = 47,22
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