Page 34 - Pra U STPM 2022 Penggal 2 - Mathematics

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 15

An object moves along a straight line and passes a fixed point O with velocity u in the positive direction of
the x-axis. At time t, the object is at a displacement x from O and the velocity of the object is v. The rate of
change of velocity has magnitude k , where k is a constant, and is directed to towards the fixed point O.
v 2
(a) Write down a differential equation for the motion of the object involving the velocity v and the time t.
Hence, find the velocity v as a function of the time t.

(b) Show that dv = v dv and hence, write down the differential equation for the motion of the object.
dx
dt
Hence, find the velocity v as function of the displacement x.
4
—
3
4
Hence, show that after a time t and the object has moved a distance x, 4kx = u – (u – 3kt) .
3
Solution: (a) The differential equation for the motion of the object involving the velocity v
as a function of the time t is dv = – k .
dt v 2
(the negative sign shows that the rate of change of velocity is directed towards
the fixed point O but the motion of the object is in the opposite direction i.e.
positive direction of the x-axis)
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Separating the variables and integrating both sides:
∫
2
∫ v dv = – kdt
v 3 = –kt + C
3
When t = 0, v = u,

4 u 3 3 = –k(0) + C ⇒ C = u 3 3

Substituting the value of C
v 3 = –kt + u 3
3 v = –3kt + u 3 3
3

3
Penerbitan (b) Using the chain rule of differentiation. dx
–3kt + u
\ v = 
3

dv
dx
dv
1 21 2

=
dx
dt
dt
The rate of change of x with respect to time t,

dv
dv
\

dt = 1 2 v dt = v
dx
dv dv
\ = v
dt dx
142

04 STPM Math(T) T2.indd 142 28/01/2022 5:44 PM