Paper 2 Kinematics Answers

Q1 Straight-line motion graph

1. Gradient represents acceleration. [1]
2. $a=\Delta v/\Delta t=(8.0-0)/4.0=2.0\mathrm{m}{\mathrm{s}}^{-2}$. [2]
3. Displacement is signed area under the graph. First triangle $=16\mathrm{m}$; rectangle $=48\mathrm{m}$; final trapezium area $=\frac{1}{2}(8.0+(-4.0))(4.0)=8.0\mathrm{m}$. Total displacement $=72\mathrm{m}$. [3]
4. The velocity becomes negative during the last interval, so the trolley reverses direction. Displacement subtracts the backward part algebraically, whereas distance adds the magnitude of all path lengths. [2]

Q2 Vertical motion

1. $a=-9.81\mathrm{m}{\mathrm{s}}^{-2}$. [1]
2. At highest point $v=0$. $0=14.0-9.81t$, so $t=1.43\mathrm{s}$. [2]
3. $v^2=u^2+2as$: $0^2=14.0^2+2(-9.81)s$, so $s=9.99\mathrm{m}$. [2]
4. Ground is at $s=-18.0\mathrm{m}$. Use $s=ut+\frac{1}{2}a{t}^2$: $-18.0=14.0t-4.905t^2$. Hence $4.905t^2-14.0t-18.0=0$, giving $t=4.11\mathrm{s}$ for the positive root. [3]
5. $v=u+at=14.0-9.81(4.11)=-26.3\mathrm{m}{\mathrm{s}}^{-1}$. The velocity is $26.3\mathrm{m}{\mathrm{s}}^{-1}$ downward. [2]

Q3 Falling with air resistance

1. At the start, speed is zero, so air resistance is negligible. Weight is the main force, so resultant force is approximately $mg$ downward and $a\approx g$. [2]
2. As speed increases, air resistance increases upward. The upward drag reduces the downward resultant force. Since $a=F_{\mathrm{result}}/m$, acceleration decreases. [3]
3. Terminal velocity occurs when air resistance equals weight, so resultant force and acceleration are zero. [1]
4. Award [1] for labelled velocity-time axes, [1] for curve starting at $v=0$ with positive initial gradient, [1] for curve flattening towards a horizontal terminal-velocity asymptote. [3]
5. Drag depends on speed, so the resultant force changes continuously as the speed changes. The acceleration tends to zero gradually, so the velocity approaches the terminal value asymptotically. [2]