Page 44 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
5 Additional Mathematics Chapter 8 Kinematics of Linear Motion

Example 20 When t = 5, s = 2(5) 3 – 6(5) + 10(5)
2
3
A particle moves along a straight line through 2
a fixed point O with velocity 10 m s . The s = –16 m
–1
3
acceleration, a m s , is given by a = 4t – 12 where Number line,
–2
t is the time, in seconds, after passing through t = 0
a fixed point O. [Assume motion to the right as
positive] Find t = 1
(a) the time, in seconds, when the acceleration is 2 0 2
zero, –16 – 3 4 – 3
(b) the minimum velocity, in m s , of the particle, t = 5
–1
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(c) the time, in seconds, when the particle stop 2 2 2
momentarily, Total distance travelled = 4 + 4 + 16 3
3
3
(d) the total distance, in m, travelled by the particle = 26
in the first 5 seconds.
Solution: Thus, in the first 5th seconds, total distance
travelled of particle is 26 m.
Given the acceleration function, a = 4t – 12
∫
∫
Velocity function, v = a dt = 4t – 12 dt Alternative Method
= 2t – 12t + c (A) Using velocity-time graph
2
Distance can be determine by finding the
When t = 0, v = 10 area under the curve.
10 = 2(0) – 12(0) + c v = 2t – 12t + 10
2
2
c = 10
So, v = 2t – 12t + 10. t 0 1 5
2
∫
Displacement function, s = v dt v 10 0 0
Velocity-time graph for 0 < t < 5:
∫
2
= 2t – 12t + 10 dt
CHAP. v
8 = 2 3 2
t – 6t + 10t + c
3
When t = 0, s = 0 10
2(0) 3
0 = 3 – 6(0) + 10(0) + c v = 2t – 12t + 10
2
2
c = 0 A
So, s = t – 6t + 10t. t
2 3
2
3 0 1 5
(a) Acceleration is zero, a = 0 B
4t – 12 = 0
4t = 12 –8
t = 3
Thus, the acceleration of the particle is zero at 1
∫
2
t = 3 seconds. Area A = 2t – 12t + 10 dt
0
(b) Minimum velocity, dv = 0, a = 0 2 1
3
2
dt = [ t – 6t + 10t ]
4t – 12 = 0 3 0
2
]
3
2
t = 3 = [ 3 (1) – 6(1) + 10(1) – 0
2
When t = 3, v = 2(3) – 12(3) + 10 2 2
= –8 = 4 m
3
Thus, the minimum velocity of the particle is ∫ 5 2
–8 m s . Area B = 2t – 12t + 10 dt
–1
1
(c) Particle stops, v = 0 2 3 2 5
2t – 12t + 10 = 0 = [ 3 t – 6t + 10t ] 1
2
(t – 1)(t – 5) = 0 = 2 (5) – 6(5) + 10(5) – 4 2
3
2
t = 1 and t = 5 [ 3 ] 3
1
Thus, the particle stops at t = 1 s and t = 5 s. = –21 
(d) When t = 0, s = 2(0) 3 – 6(0) + 10(0) 3
2
1
3 = 21 m 2
3
s = 0 m Total distance travelled = Area A + Area B
2(1) 3
When t = 1, s = – 6(1) + 10(1) 2 1
2
3 = 4 + 21 3
3
2
s = 4 m = 26 m
3
410 8.4.1
C08 Spotlight Add Math F5.indd 410 16/04/2021 5:43 PM