CAIE A-Level Mathematics 4.2 Straight-Line Kinematics Question Bank

Speed $=20 \mathrm{~ms}^{-1}$

B1

$20=0+a \times 5$

M1

Use of v=u+a t OE

$a=4 \mathrm{~ms}^{-2}$

A1

$\frac{50+100+50+200}{20}$

M1

Use of $\frac{\text { total distance }}{\text { total time }}$ OE

Average speed $=20 \mathrm{~ms}^{-1}$

A1

Attempt to differentiate v

*M1

Decrease power by 1 and a change in coefficient in at least one term (which must be the same term); allow unsimplified; allow p or q for t.
$$a=\frac{v}{t} \text { is M0. }$$$\left(a=\frac{\mathrm{d} v}{\mathrm{~d} t}=\right) 3 \times-0.1 t^{3-1}+2 \times 1.8 t^{2-1}-6 t^{1-1}=-0.3 t^{2}+3.6 t-6$

A1

May be unsimplified.

Marking guidance:

Allow p or q for t.

Setting $a=\frac{\mathrm{d} v}{\mathrm{~d} t}=0$ and attempt to solve a 3 term quadratic for t.
$$\left[a=\frac{\mathrm{d} v}{\mathrm{~d} t}=0 \Rightarrow 3 t^{2}-36 t+60=0 \Rightarrow t^{2}-12 t+20=0\right]$$

DM1

Allow p or q for t.

Must get 2 values or numerical expressions for t from their three term quadratic.

If using quadratic formula, must be the correct formula. If factorising, when brackets expanded, 2 terms correct.

p=2, q=10

A1

Velocities are $0 \mathrm{~ms}^{-1}$ and $25.6 \mathrm{~ms}^{-1}$

B1

SOI

Curve with single minimum turning point followed by single maximum turning point

*B1

Ignore placement of graph on axes
Not a cusp for the minimum point
or maximum point.

DB1

All correct in 1st and 4th quadrant.

Must go from convex to concave. Need to label 2 and 14 on the t-axis where the curve meets the t-axis.

Do not need to show exact velocities at t=0 or t=10 or t=15.

Ignore graph outside $0 \leqslant t \leqslant 15$.

Attempt to integrate v

*M1

Increase power by 1 and a change
in coefficient in at least one term (which must be the same term);
s=v t is M0.

$(s=)-\frac{0.1}{4} t^{3+1}+\frac{1.8}{3} t^{2+1}-\frac{6}{2} t^{1+1}+5.6 t(+c) =-0.025 t^{4}+0.6 t^{3}-3 t^{2}+5.6 t[+c]$$(s=)-\frac{0.1}{4} t^{3+1}+\frac{1.8}{3} t^{2+1}-\frac{6}{2} t^{1+1}+5.6 t(+c)$$=-0.025 t^{4}+0.6 t^{3}-3 t^{2}+5.6 t[+c]$

A1

May be unsimplified.

Attempt distance from t=0 to t=14[=176.4]

DM1

Correct use of limits 0 and 14 for
their s, i.e. F(14)-F(0)
May see limits 0 to 2 and 2 to 14 used but must be
(F(14)-F(2))+(F(2)-F(0)).

Attempt distance from t=14 to t=15[=(-) 8.025]

DM1

Correct use of limits 14 and 15 for
their s, i.e. $\pm(F(15)-F(14))$.
For reference $F(2)=\frac{18}{5}=3.6$,
$F(14)=\frac{882}{5}=176.4$ and
$F(15)=\frac{1347}{8}=168.375$.

Total distance $=176.4+8.025=184.425 \mathrm{~m}=\frac{7377}{40}$

A1

www
Condone 184 or better.

7(c)

SC for those who show no integration. Max 3 marks.

$\int_{0}^{14}\left(-0.1 t^{3}+1.8 t^{2}-6 t+5.6\right) \mathrm{d} t=176.4$

B1

$$\int_{14}^{15}\left(-0.1 t^{3}+1.8 t^{2}-6 t+5.6\right) \mathrm{d} t=-8.025$$
OR $\left|\int_{14}^{15}\left(-0.1 t^{3}+1.8 t^{2}-6 t+5.6\right) \mathrm{d} t\right|=8.025$

B1

Total distance $=176.4+8.025=184.425 \mathrm{~m}=\frac{7377}{40}$

B1

Condone 184 or better.

SC for those who show no integration and don't consider the 2 areas. Max 1 mark.

$\int_{0}^{15}\left|-0.1 t^{3}+1.8 t^{2}-6 t+5.6\right| \mathrm{d} t=184.425 \mathrm{~m}=\frac{7377}{40}$

B1

Condone 184 or better.