A cyclist travels along a straight road with constant acceleration. He passes through points $$\(A, B\)$$ and $$\(C\)$$. The cyclist takes 2 seconds to travel along each of the sections $$\(A B\)$$ and $$\(B C\)$$ and passes through $$\(B\)$$ with speed $$\(4.5 \mathrm{~m} \mathrm{~s}^{-1}\)$$. The distance $$\(A B\)$$ is $$\(\frac{4}{5}\)$$ of the distance $$\(B C\)$$. Find the acceleration of the cyclist. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_m20_qp_42 Year：2020 Question No：4(a)
Answer:
Use the constant acceleration equations to obtain an expression for either $s_{A B}$ or $s_{B C}$ in terms of $a$
$\begin{array}{l}s_{A B}=2 \times 4.5-1 / 2 \times a \times 2^{2} \\ s_{B C}=2 \times 4.5+1 / 2 \times a \times 2^{2}\end{array}$
$\begin{array}{l}{\left[2 \times 4.5-1 / 2 a \times 2^{2}=\frac{4}{5}\left(2 \times 4.5+1 / 2 a \times 2^{2}\right)\right]} \\ a=0.5 \mathrm{~ms}^{-2}\end{array}$
Alternative method for question 4(a)
$\left[4.5=u+2 a, s_{A C}=4 u+8 a, s_{A B}=2 u+2 a\right]$
Two correct equations
Three correct equations
$\begin{array}{l}{\left[2(4.5-2 a)+6 a=\frac{5}{4}\{2(4.5-2 a)+2 a\}\right]} \\ a=0.5 \mathrm{~ms}^{-2}\end{array}$
Alternative method for question 4 (a)
$[A C=4.5 \times 4]$
$\begin{array}{l}B C=5 / 9 \times A C \text { or } A B=4 / 9 \times A C \\ B C=10 \text { or } A B=8\end{array}$
$\begin{array}{l}{[10=4.5 \times 2+2 a \text { or } 8=4.5 \times 2-2 a]} \\ a=0.5 \mathrm{~ms}^{-2}\end{array}$
$\begin{array}{l}s_{A B}=2 \times 4.5-1 / 2 \times a \times 2^{2} \\ s_{B C}=2 \times 4.5+1 / 2 \times a \times 2^{2}\end{array}$
$\begin{array}{l}{\left[2 \times 4.5-1 / 2 a \times 2^{2}=\frac{4}{5}\left(2 \times 4.5+1 / 2 a \times 2^{2}\right)\right]} \\ a=0.5 \mathrm{~ms}^{-2}\end{array}$
Alternative method for question 4(a)
$\left[4.5=u+2 a, s_{A C}=4 u+8 a, s_{A B}=2 u+2 a\right]$
Two correct equations
Three correct equations
$\begin{array}{l}{\left[2(4.5-2 a)+6 a=\frac{5}{4}\{2(4.5-2 a)+2 a\}\right]} \\ a=0.5 \mathrm{~ms}^{-2}\end{array}$
Alternative method for question 4 (a)
$[A C=4.5 \times 4]$
$\begin{array}{l}B C=5 / 9 \times A C \text { or } A B=4 / 9 \times A C \\ B C=10 \text { or } A B=8\end{array}$
$\begin{array}{l}{[10=4.5 \times 2+2 a \text { or } 8=4.5 \times 2-2 a]} \\ a=0.5 \mathrm{~ms}^{-2}\end{array}$
Knowledge points:
4.2.4 use appropriate formulae for motion with constant acceleration in a straight line. （Questions may involve setting up more than one equation, using information about the motion of different particles.）
Solution:
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