Answer:
Acceleration
[tex]a=\left(2t-\dfrac{3}{5}\right)\:\sf ms^{-2}[/tex]
Part (A)
Velocity
[tex]\begin{aligned}\displaystyle v= \int a\:dt & =\int \left(2t-\dfrac{3}{5}\right)\:dt\\\\ & = t^2 -\dfrac{3}{5}t+C \end{aligned}[/tex]
[tex]\begin{aligned} \textsf{As }\:v(0)=10 \implies (0)^2-\dfrac{3}{5}(0)+C & =10\\C & = 10 \end{aligned}[/tex]
[tex]\implies v(t)=\left(t^2-\dfrac{3}{5}t+10\right)\: \sf ms^{-1}[/tex]
Part (B)
Displacement
[tex]\begin{aligned}\displaystyle s= \int v\:dt & =\int \left(t^2-\dfrac{3}{5}t+10\right)\:dt\\\\ & = \dfrac{1}{3}t^3-\dfrac{3}{10}t^2+10t+C \end{aligned}[/tex]
[tex]\begin{aligned} \textsf{As }\: s(0)=5 \implies \dfrac{1}{3}(0)^3-\dfrac{3}{10}(0)^2+10(0)+C & =5\\C & = 5\end{aligned}[/tex]
[tex]\implies s(t)=\left(\dfrac{1}{3}t^3-\dfrac{3}{10}t^2+10t+5\right)\: \sf m[/tex]
Part (C)
(i) t = 2 s
[tex]v(2)=(2)^2-\dfrac{3}{5}(2)+10=\dfrac{64}{5}=12.8\: \sf ms^{-1}[/tex]
[tex]s(2)=\dfrac{1}{3}(2)^3-\dfrac{3}{10}(2)^2+10(2)+5=\dfrac{397}{15}=26.47\: \sf m\:(2\:dp)[/tex]
(ii) t = 5 s
[tex]v(5)=(5)^2-\dfrac{3}{5}(5)+10=32\: \sf ms^{-1}[/tex]
[tex]s(5)=\dfrac{1}{3}(5)^3-\dfrac{3}{10}(5)^2+10(5)+5=\dfrac{535}{6}=89.17\: \sf m\:(1\:dp)[/tex]
