Paper 3 Kinematics Answers

Q1 Non-uniform motion from a graph

1. The acceleration is initially positive because velocity increases. The increase in velocity per second becomes smaller, so the positive acceleration decreases. After about $t=5\mathrm{s}$, velocity decreases, so acceleration becomes negative. [3]
2. Use trapezium rule with interval $1\mathrm{s}$:

$$s\approx \frac{1}{2}(0+2(0.80+1.45+1.92+2.18+2.25+2.13+1.82)+1.35).$$

This gives $s\approx 13.2\mathrm{m}$. [4] 3. Use a tangent or central difference near $t=4\mathrm{s}$:

$$a\approx \frac{v(5)-v(3)}{5-3}=\frac{2.25-1.92}{2}=0.165\mathrm{m}{\mathrm{s}}^{-2}.$$

Accept nearby values from a sensible tangent method. [4] 4. Uniform acceleration would give a straight-line velocity-time graph with constant gradient. The data show changing gradient, including a change from positive to negative acceleration, so one constant acceleration cannot represent the whole interval. [3]

Q2 Constant-acceleration model and sign conventions

1. $v=u+at=0$: $0=1.20-0.480t$, so $t=2.50\mathrm{s}$. [2]
2. $s=ut+\frac{1}{2}a{t}^2=1.20(6.00)+\frac{1}{2}(-0.480)(6.00{)}^2=-1.44\mathrm{m}$. [3]
3. Forward displacement before stopping:

$$s_1=\frac{u^2}{2|a|}=\frac{1.20^2}{2(0.480)}=1.50\mathrm{m}.$$

Remaining time $=3.50\mathrm{s}$ from rest with acceleration in the negative direction:

$$s_2=\frac{1}{2}(0.480)(3.50{)}^2=2.94\mathrm{m}.$$

Total distance $=1.50+2.94=4.44\mathrm{m}$. [4] 4. The glider reverses direction after $2.50\mathrm{s}$. Displacement is signed, so the backward motion subtracts from the forward displacement; distance adds both path lengths. [2] 5. Award [1] for linear graph with negative gradient, [1] for crossing $v=0$ at $2.50\mathrm{s}$, [1] for continuing to negative velocity at $6.00\mathrm{s}$. [3]