IB Maths AA HL 5.9 Kinematics Question Bank

valid approach to find turning point $\left(v^{\prime}=0,-\frac{b}{2 a}\right.$, average of roots)
4-6 t=0 OR $-\frac{4}{2(-3)}$ OR $\frac{-\frac{2}{3}+2}{2}$$t=\frac{2}{3}(\mathrm{~s})$

attempt to integrate v
$\int v \mathrm{~d} t=\int\left(4+4 t-3 t^{2}\right) \mathrm{d} t=4 t+2 t^{2}-t^{3}(+c)$
Note: Award A1 for $4 t+2 t^{2}, \boldsymbol{A} \mathbf{1}$ for $-t^{3}$.
attempt to substitute their t into their solution for the integral
distance $=4\left(\frac{2}{3}\right)+2\left(\frac{2}{3}\right)^{2}-\left(\frac{2}{3}\right)^{3}$$=\frac{8}{3}+\frac{8}{9}-\frac{8}{27}$ (or equivalent)
$=\frac{88}{27}(\mathrm{~m})$

recognising to integrate between 0 and 2, or 2 and 3 OR $\int_{0}^{3}\left|4+4 t-3 t^{2}\right| \mathrm{d} t$$\int_{0}^{2}\left(4+4 t-3 t^{2}\right) d t$
=8
A1
$\int_{2}^{3}\left(4+4 t-3 t^{2}\right) d t$
=-5
A1
valid approach to sum the two areas (seen anywhere)
$\int_{0}^{2} v \mathrm{~d} t-\int_{2}^{3} v \mathrm{~d} t \quad$ OR $\int_{0}^{2} v \mathrm{~d} t+\left|\int_{2}^{3} v \mathrm{~d} t\right|$
total distance travelled $=13(\mathrm{~m})$