IB Maths AA HL Sl 5 9 Kinematics Questions

Practise identifying the mathematical structure in a kinematics problem, selecting the correct theorem or algorithm, and checking the result with units, graphs or logical…

Exam points

Differentiate and integrate position, displacement and velocity functions to obtain velocity, acceleration and displacement in one-dimensional motion.
Determine displacement, distance travelled and speed over a time interval, using sign changes and integrals of velocity where required.
Solve and interpret rest, change-of-direction, maximum-speed and zero-acceleration conditions for particles, including motion in context.
Use advanced kinematic relationships, including acceleration as v dv/ds and exact integration, to solve particle-motion problems.

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Study SL 5.9—Kinematics conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

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A particle P moves along the x-axis. The velocity of P is $v \mathrm{~m} \mathrm{~s}^{-1}$ at time t seconds, where $v(t)=4+4 t-3 t^{2}$ for $0 \leq t \leq 3$ . When t=0, P is at the origin O .

(a)

Find the value of t when P reaches its maximum velocity.

[ 2 ]

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 $\frac{-\frac{2}{3}+2}{2}$ $t=\frac{2}{3}(\mathrm{~s})$

(b)

Show that the distance of P from O at this time is $\frac{88}{27}$ metres.

[ 7 ]

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)
$=\frac{88}{27}(\mathrm{~m})$

(c)

Find the total distance travelled by P.

[ 5 ]

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
$\int_{2}^{3}\left(4+4 t-3 t^{2}\right) d t$
=-5
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 $=13(\mathrm{~m})$

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IB Maths AA HL Sl 5 9 Kinematics Questions question 1

Question (a)

Question (b)

Question (c)