Page 43 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
5 Additional Mathematics Chapter 8 Kinematics of Linear Motion

2. The diagram below shows the movement of (c) Displacement is maximum when v = 0, which is
an object thrown by Ahmad and its value of t = 4 s.
displacement and velocity with respect to the time. When t = 4, s = 4 + 8(4) – (4) 2
s = 4 + 32 – 16
t = 2 s s = 20 m
s = 20 m  Maximum Thus, the maximum displacement of the
v = 0 displacement
particle is 20 m.
(d) Velocity decreases, v , 0
t = 1 s t = 3 s
s = 15 m   s = 15 m 8 – 2t , 0
v = +10 m/s v = –10 m/s 2t . 8
©PAN ASIA PUBLICATIONS
t . 4
Initial velocity
Thus, the range of value t is t . 4 s.
t = 0 (e) When the particle moves to the right, v . 0
s = 0  t = 4 s 8 – 2t . 0
v = +20 m/s  s = 0 2t , 8
v = –20 m/s
t , 4
Thus, the time interval of the particle moves to
the right is t , 4 s.
Try question 4 in Formative Zone 8.2
• Particles stop, so velocity, v = 0
• For maximum or minimum displacement,
ds = v = 0 Example 12
dt A particle moves along a straight line with its
• Particles return or through a fixed point O, displacement, s meters, from a fixed point O given
s = 0 by s = t – 12t + 36t + 10, where t is the time, in
2
3
CHAP. • Particles change their direction of movement, seconds. [Assume motion to the right as positive]
8 v = 0 Find –1
(a) the initial velocity, in m s , of the particle,
(b) the values of t, in seconds, when the particle
Example 11 stops instantaneously,
(c) the range of value t when the velocity is
A particle moves along a straight line with its negative.
displacement, s meters, from a fixed point O given
by s = 4 + 8t – t , where t is the time, in seconds. Solution:
2
[Assume motion to the right as positive] Given s = t – 12t + 36t + 10.
2
3
Find Velocity function, v = ds = 3t – 24t + 36
2
(a) the initial velocity, in m s , of the particle, dt
–1
2
(b) the value of t, in seconds, when the particle (a) Initial velocity, t = 0 v = 3(0) – 24(0) + 36
stops instantaneously, = 36
(c) the maximum displacement of the particle, Thus, the initial velocity of the particle is 36 m s .
–1
(d) the range of values t values when the velocity (b) Particle stops, v = 0
decreases, 3t – 24t + 36 = 0
2
(e) the time interval when the particle moves to (t – 6)(t – 2) = 0
the right. t = 6 or t = 2
Thus, the particle stops at t = 6 s and t = 2 s.
Solution: (c) Velocity is negative, v , 0
Given s = 4 + 8t – t . 3t – 24t + 36 , 0
2
2
Velocity function, v = ds = 8 – 2t (t – 6)(t – 2) , 0
dt Let (t – 6)(t – 2) = 0
(a) Initial velocity, t = 0 t = 6 or 2
When t = 0, v = 8 – 2(0)
= 8
–1
Thus, the intial velocity of the particle is 8 m s .
(b) Particle stops, v = 0 t
8 – 2t = 0 2 6
2t = 8 Thus, the range of value t is 2 , t , 6 s.
t = 4
Thus, the particle stops at t = 4 s. Try question 5 in Formative Zone 8.2
404 8.2.2

C08 Spotlight Add Math F5.indd 404 16/04/2021 5:43 PM