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 D = 0

x =

√ 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 terms 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 250 will cool from 100

0 to 75

0 in

one minute, find its temperature at the end off three minutes

Solution : let temperature of the body be T0 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)

= e

k

(1) becomes T = 25 + 75 ekt

When t = 3 then T = 25 + 75 ekt

= 25 + 75 x 8/27 = 47,22