Soalan 17:
Rajah 6 menunjukkan graf laju-masa bagi dua orang pelari. Graf OP mewakili larian Ahmad dan graf OQ mewakili larian Goh.
Hitung beza antara pecutan, dalam m s-2, bagi larian mereka.
A 0.2
B 0.3
C 0.4
D 0.6
Penyelesaian:
Jawapan : C
$$ \begin{aligned} &\text { Pecutan larian Ahmad }\\ &\begin{aligned} & =\frac{6 \mathrm{~m} \mathrm{~s}^{-1}}{10 \mathrm{~s}} \\ & =0.6 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \end{aligned} $$
$$ \begin{aligned} \text { Pecutan larian Goh } & =\frac{6 \mathrm{~m} \mathrm{~s}^{-1}}{30 \mathrm{~s}} \\ & =0.2 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} &\text { Beza antara pecutan }\\ &\begin{aligned} & =0.6 \mathrm{~m} \mathrm{~s}^{-2}-0.2 \mathrm{~m} \mathrm{~s}^{-2} \\ & =0.4 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \end{aligned} $$
Rajah 6 menunjukkan graf laju-masa bagi dua orang pelari. Graf OP mewakili larian Ahmad dan graf OQ mewakili larian Goh.
Hitung beza antara pecutan, dalam m s-2, bagi larian mereka.
A 0.2
B 0.3
C 0.4
D 0.6
Penyelesaian:
Jawapan : C
$$ \begin{aligned} &\text { Pecutan larian Ahmad }\\ &\begin{aligned} & =\frac{6 \mathrm{~m} \mathrm{~s}^{-1}}{10 \mathrm{~s}} \\ & =0.6 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \end{aligned} $$
$$ \begin{aligned} \text { Pecutan larian Goh } & =\frac{6 \mathrm{~m} \mathrm{~s}^{-1}}{30 \mathrm{~s}} \\ & =0.2 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} &\text { Beza antara pecutan }\\ &\begin{aligned} & =0.6 \mathrm{~m} \mathrm{~s}^{-2}-0.2 \mathrm{~m} \mathrm{~s}^{-2} \\ & =0.4 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} \end{aligned} $$
Soalan 18:
$$ \text { Diberi bahawa } m\left[\begin{array}{l} n \\ 8 \end{array}\right]=\left[\begin{array}{c} 20 \\ -40 \end{array}\right] $$
$$ \text { Cari nilai } m \text { dan nilai } n \text {. } $$
$$ \begin{aligned} &\text { A } m=-5, n=-4\\ &\text { B } m=-5, n=4\\ &\text { C } m=-4, n=-5\\ &\text { D } m=-4, n=5 \end{aligned} $$
Penyelesaian:
Jawapan : A
$$ \begin{aligned} m\left[\begin{array}{l} n \\ 8 \end{array}\right] & =\left[\begin{array}{c} 20 \\ -40 \end{array}\right] \\ m(8) & =-40 \\ m & =-\frac{40}{8} \\ m & =-5 \end{aligned} $$
$$ \begin{aligned} m n & =20 \\ (-5) n & =20 \\ n & =-\frac{20}{5} \\ n & =-4 \end{aligned} $$
$$ \text { Diberi bahawa } m\left[\begin{array}{l} n \\ 8 \end{array}\right]=\left[\begin{array}{c} 20 \\ -40 \end{array}\right] $$
$$ \text { Cari nilai } m \text { dan nilai } n \text {. } $$
$$ \begin{aligned} &\text { A } m=-5, n=-4\\ &\text { B } m=-5, n=4\\ &\text { C } m=-4, n=-5\\ &\text { D } m=-4, n=5 \end{aligned} $$
Penyelesaian:
Jawapan : A
$$ \begin{aligned} m\left[\begin{array}{l} n \\ 8 \end{array}\right] & =\left[\begin{array}{c} 20 \\ -40 \end{array}\right] \\ m(8) & =-40 \\ m & =-\frac{40}{8} \\ m & =-5 \end{aligned} $$
$$ \begin{aligned} m n & =20 \\ (-5) n & =20 \\ n & =-\frac{20}{5} \\ n & =-4 \end{aligned} $$
Soalan 19:
$$ \text { Diberi bahawa } 3\left[\begin{array}{lll} -1 & 2 & 5 \end{array}\right]-\frac{1}{2}\left[\begin{array}{lll} 3 & 4 & p \end{array}\right]=\left[\begin{array}{lll} t & k & 0 \end{array}\right] \text {, } $$
$$ \text { cari nilai } p \text {, nilai } t \text { dan nilai } k \text {. } $$
Penyelesaian:
Jawapan : A
$$ \begin{aligned} & 3\left[\begin{array}{lll} -1 & 2 & 5 \end{array}\right]-\frac{1}{2}\left[\begin{array}{lll} 3 & 4 & p \end{array}\right]=\left[\begin{array}{lll} t & k & 0 \end{array}\right] \\ & {\left[\begin{array}{lll} -3 & 6 & 15 \end{array}\right]-\left[\begin{array}{lll} \frac{3}{2} & \frac{4}{2} & \frac{p}{2} \end{array}\right]=\left[\begin{array}{lll} t & k & 0 \end{array}\right]} \end{aligned} $$
$$ \begin{aligned} -3-\frac{3}{2} & =t \\ t & =-\frac{9}{2} \end{aligned} $$
$$ \begin{aligned} 6-\frac{4}{2} & =k \\ k & =4 \end{aligned} $$
$$ \begin{aligned} 15-\frac{p}{2} & =0 \\ 15 & =\frac{p}{2} \\ p & =30 \end{aligned} $$
$$ \text { Diberi bahawa } 3\left[\begin{array}{lll} -1 & 2 & 5 \end{array}\right]-\frac{1}{2}\left[\begin{array}{lll} 3 & 4 & p \end{array}\right]=\left[\begin{array}{lll} t & k & 0 \end{array}\right] \text {, } $$
$$ \text { cari nilai } p \text {, nilai } t \text { dan nilai } k \text {. } $$
Penyelesaian:
Jawapan : A
$$ \begin{aligned} & 3\left[\begin{array}{lll} -1 & 2 & 5 \end{array}\right]-\frac{1}{2}\left[\begin{array}{lll} 3 & 4 & p \end{array}\right]=\left[\begin{array}{lll} t & k & 0 \end{array}\right] \\ & {\left[\begin{array}{lll} -3 & 6 & 15 \end{array}\right]-\left[\begin{array}{lll} \frac{3}{2} & \frac{4}{2} & \frac{p}{2} \end{array}\right]=\left[\begin{array}{lll} t & k & 0 \end{array}\right]} \end{aligned} $$
$$ \begin{aligned} -3-\frac{3}{2} & =t \\ t & =-\frac{9}{2} \end{aligned} $$
$$ \begin{aligned} 6-\frac{4}{2} & =k \\ k & =4 \end{aligned} $$
$$ \begin{aligned} 15-\frac{p}{2} & =0 \\ 15 & =\frac{p}{2} \\ p & =30 \end{aligned} $$
Soalan 20:
Diberi bahawa p berubah secara songsang dengan kuasa dua w dan p = 3 apabila w = 2.
Cari hubungan antara p dan w.
$$ \begin{aligned} &\text { A } p=12 w\\ &\text { B } p=\frac{12}{w^2}\\ &\text { C } p=\frac{3}{4} w^2\\ &\text { D } p=\frac{3}{4 w^2} \end{aligned} $$
Penyelesaian:
Jawapan : B
$$ \begin{aligned} & p ∝ \frac{1}{w^2} \\ & p=\frac{k}{w^2} \\ & 3=\frac{k}{2^2} \\ & k=3 \times 2^2 \\ & k=12 \\ & p=\frac{12}{w^2} \end{aligned} $$
Diberi bahawa p berubah secara songsang dengan kuasa dua w dan p = 3 apabila w = 2.
Cari hubungan antara p dan w.
$$ \begin{aligned} &\text { A } p=12 w\\ &\text { B } p=\frac{12}{w^2}\\ &\text { C } p=\frac{3}{4} w^2\\ &\text { D } p=\frac{3}{4 w^2} \end{aligned} $$
Penyelesaian:
Jawapan : B
$$ \begin{aligned} & p ∝ \frac{1}{w^2} \\ & p=\frac{k}{w^2} \\ & 3=\frac{k}{2^2} \\ & k=3 \times 2^2 \\ & k=12 \\ & p=\frac{12}{w^2} \end{aligned} $$
Soalan 21:
Jadual 5 menunjukkan beberapa nilai pemboleh ubah R, S dan T.
Diberi bahawa R berubah secara langsung dengan S dan secara songsang dengan T.
Hitung nilai y apabila p × q = 20.
$$ \begin{aligned} &\text { A } \frac{1}{48}\\ &\text { B } \frac{1}{3}\\ &\text { C } 3\\ &\text { D } 48 \end{aligned} $$
Penyelesaian:
Jawapan : B
$$ \begin{aligned} R ∝ \frac{S}{T} \\ p & =\frac{k S}{T} \\ p & =\frac{k(5)}{q} \\ p \times q & =5 k \\ 20 & =5 k \\ k & =\frac{20}{5} \\ k & =4 \end{aligned} $$
$$ \begin{aligned} y & =\frac{4\left(\frac{1}{2}\right)}{6} \\ y & =\frac{1}{3} \end{aligned} $$
Jadual 5 menunjukkan beberapa nilai pemboleh ubah R, S dan T.
Diberi bahawa R berubah secara langsung dengan S dan secara songsang dengan T.Hitung nilai y apabila p × q = 20.
$$ \begin{aligned} &\text { A } \frac{1}{48}\\ &\text { B } \frac{1}{3}\\ &\text { C } 3\\ &\text { D } 48 \end{aligned} $$
Penyelesaian:
Jawapan : B
$$ \begin{aligned} R ∝ \frac{S}{T} \\ p & =\frac{k S}{T} \\ p & =\frac{k(5)}{q} \\ p \times q & =5 k \\ 20 & =5 k \\ k & =\frac{20}{5} \\ k & =4 \end{aligned} $$
$$ \begin{aligned} y & =\frac{4\left(\frac{1}{2}\right)}{6} \\ y & =\frac{1}{3} \end{aligned} $$
Soalan 22:
Rajah 7 menunjukkan sebuah prisma yang terletak pada satah mengufuk. RSTW ialah keratan rentas seragam prisma tersebut.
Antara berikut, manakah unjuran ortogon yang betul sebagaimana dilihat dari arah X?
Penyelesaian:
Jawapan : D
Rajah 7 menunjukkan sebuah prisma yang terletak pada satah mengufuk. RSTW ialah keratan rentas seragam prisma tersebut.
Antara berikut, manakah unjuran ortogon yang betul sebagaimana dilihat dari arah X?
Penyelesaian:
Jawapan : D