Kinematics Graphs: past paper questions & worked solutions

The velocity becomes zero at $t=2.5$ s. Since acceleration is $-g=-9.8\ \mathrm{m\,s^{-2}}$, the initial velocity is $u=9.8(2.5)$.

$u=24.5\ \mathrm{m\,s^{-1}}$.

From launch to $t=7$, the displacement is the area under the velocity-time graph. Algebraically, $s=ut-\dfrac12gt^2$ with $t=7$.

$s=24.5(7)-\dfrac12(9.8)(7^2)=171.5-240.1=-68.6$ m.

The negative sign shows the ground is below A, so the height of A above the ground is $68.6\ \mathrm{m}$.

At the highest point B, the velocity is $0$ and $AB=19.6$ m. Use $v^2=u^2+2as$ upward: $0=u^2-2g(19.6)$.

$u^2=2(9.8)(19.6)=384.16$, so $u=19.6\ \mathrm{m\,s^{-1}}$.

From A to the ground, taking upward as positive, displacement is $-24.5$ m. Use $s=ut+\dfrac12at^2$: $-24.5=19.6T-4.9T^2$.

$4.9T^2-19.6T-24.5=0$. Divide by $4.9$: $T^2-4T-5=0$, so $(T-5)(T+1)=0$. The positive time is $T=5$ s.

The speed starts at $19.6$, decreases linearly to $0$ at $t=2$, then increases linearly until the ball hits the ground at $t=5$. The final speed is $29.4$, so the second line is longer than the first.

The speed is less than $14.7\ \mathrm{m\,s^{-1}}$ near the highest point. From speed $14.7$ to speed $0$, use $v^2=u^2+2as$: $0=14.7^2-2gs$.

This gives $s=11.025$ m on the way up. The same distance occurs on the way down before the speed reaches $14.7$ again, so the total distance is $2s=22.05\ \mathrm{m}$, about $22.1\ \mathrm{m}$.

The ball has speed $29.4$ again when it passes A on the way down. It is faster than $29.4$ only after passing A downward until it hits the ground.

From A to the ground, taking downward as positive, $19.6=29.4t+\dfrac12gt^2$.

$19.6=29.4t+4.9t^2$, giving the positive root $t=0.606\ldots\ \mathrm{s}$.

The graph starts at speed $29.4$ when $t=0$, decreases linearly to speed $0$ at $t=3$, then increases linearly until the ball hits the ground. The final speed is greater than $29.4$, so the right-hand arm ends above the starting height on the speed axis.

Distance is the area under a speed-time graph. The three areas are a triangle from $0$ to $12$, a rectangle from $12$ to $31$, and a trapezium from $31$ to $41$.

$\dfrac12(12)(3U)+19(3U)+\dfrac12(3U+2U)(10)=600$. This gives $18U+57U+25U=600$, so $100U=600$ and $U=6$.

Since $U=6$, the final speed $6\ \mathrm{m\,s^{-1}}$ occurs during the first section. The acceleration there is $\dfrac{18-0}{12}=1.5\ \mathrm{m\,s^{-2}}$, so the time to reach $6$ is $4$ s. The distance is $\dfrac12(4)(6)=12\ \mathrm{m}$.

From $0$ to $12$, $a=\dfrac{18}{12}=1.5\ \mathrm{m\,s^{-2}}$. From $31$ to $41$, the speed falls from $18$ to $12$, so the deceleration is $\dfrac{18-12}{10}=0.6\ \mathrm{m\,s^{-2}}$.

The acceleration-time graph has three horizontal sections: $1.5$ from $0\leq t\leq 12$, $0$ from $12<t\leq31$, and $-0.6$ from $31<t\leq41$.

(i) Acceleration for $0 \leq t \leq 10$ $\text{Acceleration} = \frac{\Delta v}{\Delta t} = \text{gradient of v-t graph}$ $a = \frac{30 - 0}{10 - 0}$ $a = 3 \text{ m s}^{-2}$ (ii) Deceleration for $30 \leq t \leq 35$ $a = \frac{0 - 30}{35 - 30}$ $a = \frac{-30}{5}$ $a = -6 \text{ m s}^{-2}$ Deceleration = $|-6| = 6$ m s$^{-2}$

Analysis of each section: ; $0 \leq t \leq 10$: Constant acceleration = 3 m s$^{-2}$ ; $10 < t \leq 30$: Constant velocity, so acceleration = 0 m s$^{-2}$ ; $30 < t \leq 35$: Constant deceleration = $-6$ m s$^{-2}$

$\text{Distance} = \text{Area under v-t graph}$ The shape is a trapezium. We can split it into regions: Method 1: Trapezium formula $\text{Area} = \frac{1}{2}(a + b)h$ Where the parallel sides are the time intervals and height is velocity. Better to calculate as sum of shapes: Method 2: Sum of triangle + rectangle + triangle $\text{Area}_1 = \frac{1}{2} \times 10 \times 30 = 150 \text{ m}$ $\text{Area}_2 = 20 \times 30 = 600 \text{ m}$ $\text{Area}_3 = \frac{1}{2} \times 5 \times 30 = 75 \text{ m}$ $\text{Total distance} = 150 + 600 + 75$ $= 825 \text{ m}$

Area under v-t graph from $t = 0$ to $t = 20$: $\text{Distance}_{\text{car}} = \text{Triangle} + \text{Rectangle}$ $= \frac{1}{2}(10)(30) + (10)(30)$ $= 150 + 300$ $= 450 \text{ m}$

Motorcycle starts at $t = 5$ with constant acceleration $A$. Time of travel: $20 - 5 = 15$ seconds $s = ut + \frac{1}{2}at^2$ $\text{Distance}_{\text{motorcycle}} = 0 + \frac{1}{2}A(15)^2$ $= \frac{225A}{2}$ $= 112.5A \text{ m}$

Motorcycle catches up with car when $t = 20$: $112.5A = 450$ $A = \frac{450}{112.5}$ $A = 4 \text{ m s}^{-2}$

From $0$ to $4$ s the acceleration is $-0.5\ \mathrm{m\,s^{-2}}$. Therefore $w=8+(-0.5)(4)=6\ \mathrm{m\,s^{-1}}$.

From $10$ to $20$ s the acceleration is $1.2\ \mathrm{m\,s^{-2}}$, while the speed at $t=10$ is still $6$. Hence $v=6+1.2(10)=18\ \mathrm{m\,s^{-1}}$.

The graph starts at $(0,8)$, falls linearly to $(4,6)$, stays horizontal to $(10,6)$, then rises linearly to $(20,18)$.

The distance is the area under the graph. From $0$ to $4$: $\dfrac{8+6}{2}\cdot4=28$. From $4$ to $10$: $6\cdot6=36$. From $10$ to $20$: $\dfrac{6+18}{2}\cdot10=120$.

$AB=28+36+120=184\ \mathrm{m}$.

The train reaches $20\ \mathrm{m\,s^{-1}}$ in $T$ seconds. Since the acceleration magnitude is twice the deceleration magnitude, the time needed to decelerate from $20$ to $0$ is $2T$.

The speed-time graph rises linearly from $(0,0)$ to $(T,20)$, stays horizontal at speed $20$ from $T$ to $T+180$, then falls linearly to zero at $3T+180$.

The distance from A to B is $4.8\ \mathrm{km}=4800\ \mathrm{m}$. The area is $\dfrac12(T)(20)+20(180)+\dfrac12(2T)(20)$.

$4800=10T+3600+20T$, so $1200=30T$ and $T=40\ \mathrm{s}$.

During the first $T$ seconds, $20=aT$. With $T=40$, $a=\dfrac{20}{40}=0.5\ \mathrm{m\,s^{-2}}$.

The car's graph is a horizontal line at $40\ \mathrm{m\,s^{-1}}$. The motorcyclist's graph starts at zero, rises with gradient $4$ until $t=15$, then rises with smaller gradient $1$ for a further $T$ seconds.

Before $t=15$, the motorcyclist's speed is $4t$. Set this equal to the car's speed: $4t_1=40$, so $t_1=10$ s.

After the first $15$ seconds, the motorcyclist's speed is $4(15)=60\ \mathrm{m\,s^{-1}}$.

During the next $T$ seconds the acceleration is $1\ \mathrm{m\,s^{-2}}$, so the speed at catch-up is $60+T$.

Motorcyclist distance $=\dfrac12(15)(60)+\dfrac12[60+(60+T)]T$. Car distance $=40(15+T)$.

$450+60T+\dfrac12T^2=600+40T$. Hence $\dfrac12T^2+20T-150=0$, so $T^2+40T-300=0$ and $k=40$.

The graph starts at speed $5$ when $t=0$, rises linearly to speed $V$ at time $T_1$, then falls linearly to speed $0$ at time $T_1+T_2$.

The area under the first section is the distance $AB=132$. Thus $\dfrac12(5+V)T_1=132$, so $T_1=\dfrac{264}{V+5}$.

The second section is a triangle of height $V$ and base $T_2$. Since $BC=136$, $\dfrac12VT_2=136$, so $T_2=\dfrac{272}{V}$.

The whole journey takes $28$ s, so $\dfrac{264}{V+5}+\dfrac{272}{V}=28$.

Multiplying by $V(V+5)$ gives $264V+272(V+5)=28V(V+5)$. This simplifies to $7V^2-99V-340=0$, so $(V-17)(7V+20)=0$. Since speed is positive, $V=17$.

For the final section, $T_2=\dfrac{272}{17}=16$ s. The speed falls from $17$ to $0$, so the deceleration is $\dfrac{17}{16}=1.0625\ \mathrm{m\,s^{-2}}$.

The car graph is a triangle from $0$ to $30$ plus a rectangle from $30$ to $T$. Distance $=\dfrac12(30)V+V(T-30)$.

$\dfrac12(30)V+V(T-30)=15V+VT-30V=V(T-15)$, as required.

The motorbike starts at $(10,0)$, accelerates with the same gradient as the car until $t=50$, then travels at constant speed until $t=60$, then decelerates linearly to $(T,V)$.

The car acceleration is $\dfrac{V}{30}$. The motorbike accelerates for $40$ s, so its speed is $40\cdot\dfrac{V}{30}=\dfrac{4V}{3}$.

Its distance is $\dfrac12(40)\left(\dfrac{4V}{3}\right)+10\left(\dfrac{4V}{3}\right)+\dfrac12\left(\dfrac{4V}{3}+V\right)(T-60)$.

This becomes $\dfrac{80V}{3}+\dfrac{40V}{3}+\dfrac{7V}{6}(T-60)=40V+\dfrac{7V}{6}(T-60)=\dfrac{7VT}{6}-30V$.

At time $T$, the distances are equal: $V(T-15)=\dfrac{7VT}{6}-30V$. Divide by $V$ to get $T-15=\dfrac{7T}{6}-30$.

Multiplying by $6$ gives $6T-90=7T-180$, so $T=90$.

Take upward as positive from A. At B, the displacement is $-2.5$ m, with $u=24$ and $a=-9.8$. Use $v^2=u^2+2as$: $V^2=24^2+2(-9.8)(-2.5)=625$.

$V=25\ \mathrm{m\,s^{-1}}$.

The velocity at B is $-25\ \mathrm{m\,s^{-1}}$. Using $v=u+at$, $-25=24-9.8t$, so $t=5\ \mathrm{s}$.

C is $10$ m above A. The particle is above C between the two times when $s=10$: $10=24t-4.9t^2$.

$4.9t^2-24t+10=0$, giving $t=0.459\ldots$ and $t=4.438\ldots$.

The length of time above C is $4.438\ldots-0.459\ldots=3.98\ldots\ \mathrm{s}$, about $4.0\ \mathrm{s}$.

The graph starts at speed $24$, decreases linearly to $0$ at the highest point, then increases linearly to speed $25$ at $t=5$. It is a V-shaped speed-time graph.

The parachutist falls freely from rest for $2.5$ s, so $v=gt=9.8(2.5)=24.5\ \mathrm{m\,s^{-1}}$.

The speed-time graph starts at $0$, rises linearly to $24.5$ at $t=2.5$, then decreases linearly with smaller gradient for $T$ seconds, then stays horizontal at speed $V$ until $t=20$.

During the deceleration phase, initial speed is $24.5$, acceleration is $-3.9$, time is $T$, and distance is $73.75$.

$73.75=24.5T+\dfrac12(-3.9)T^2=24.5T-1.95T^2$.

Solving gives $T=5$ or $T=7.56\ldots$. Since $T<6$, take $T=5$.

$V=24.5-3.9(5)=5\ \mathrm{m\,s^{-1}}$.

Height is the total area under the speed-time graph: first triangle $\dfrac12(2.5)(24.5)=30.625$, deceleration distance $73.75$, and final rectangle $(20-2.5-5)(5)=62.5$.

Total height $=30.625+73.75+62.5=166.875$ m, which rounds to $167$ m.

The graph starts at $(0,0)$, rises in a straight line with gradient $0.8$ until time $T$ and speed $V$, then stays horizontal at speed $V$ until $t=30$.

Uniform acceleration gives $V=0.8T$.

The distance is the area under the speed-time graph: $\dfrac12TV+V(30-T)=200$.

Using $V=0.8T$: $\dfrac12T(0.8T)+(0.8T)(30-T)=200$.

$0.4T^2+24T-0.8T^2=200$, so $-0.4T^2+24T-200=0$. Multiplying by $-2.5$ gives $T^2-60T+500=0$, so $k=60$.

$T^2-60T+500=(T-10)(T-50)=0$, so $T=10$ or $50$. Since the whole race lasts $30$ s, $T=10$.

A more realistic model could include reaction time at the start, and could replace the instantaneous change from acceleration to constant speed with a smoother transition.

For cyclist P, the distance from A to B is the area under the speed-time graph from $0$ to $42$. This is $\dfrac12(9)V+33V=120$.

$4.5V+33V=37.5V=120$, so $V=3.2\ \mathrm{m\,s^{-1}}$.

Between $t=0$ and $t=9$, P accelerates from $0$ to $3.2$. Hence $a=\dfrac{3.2}{9}=\dfrac{16}{45}=0.356\ldots\ \mathrm{m\,s^{-2}}$.

Cyclist Q starts at $t=6$ from rest, accelerates in a straight line for $T$ seconds to speed $3.6\ \mathrm{m\,s^{-1}}$, then continues horizontally at $3.6\ \mathrm{m\,s^{-1}}$ until $t=54$.

By $t=54$, P has travelled $\dfrac12(9)(3.2)+45(3.2)=14.4+144=158.4\ \mathrm{m}$.

Q travels for $48$ seconds after starting at $t=6$. Her distance is $\dfrac12T(3.6)+(48-T)(3.6)=172.8-1.8T$.

Q catches P at $t=54$, so $172.8-1.8T=158.4$. Thus $1.8T=14.4$ and $T=8$ s.

Ajaz starts from rest, accelerates linearly to speed $4\ \mathrm{m\,s^{-1}}$, then continues with a horizontal speed-time line at speed $4$.

Beth starts from rest, has a steeper straight acceleration line for $T$ seconds, then a horizontal line. Both graphs finish at the same final time.

Ajaz reaches speed $4$ with acceleration $0.8$, so the time is $\dfrac{4}{0.8}=5$ s.

Distance during acceleration is $\dfrac12(5)(4)=10$ m. The remaining $90$ m at speed $4$ takes $22.5$ s, so Ajaz's total time is $27.5$ s.

Beth accelerates for $T$ seconds to speed $T$, then runs at speed $T$ until time $27.5$. Her distance is $\dfrac12T^2+T(27.5-T)$.

$100=\dfrac12T^2+27.5T-T^2=27.5T-\dfrac12T^2$. Thus $T^2-55T+200=0$. The valid root is $T=3.915\ldots$ s.

Ajaz finishes at $4\ \mathrm{m\,s^{-1}}$. Beth's constant speed is $T=3.915\ldots\ \mathrm{m\,s^{-1}}$.

Difference $=4-3.915\ldots=0.085\ \mathrm{m\,s^{-1}}$ approximately.

Tram A starts at $t=0$, accelerates at $2\ \mathrm{m\,s^{-2}}$ for $5$ s, reaching speed $10\ \mathrm{m\,s^{-1}}$, then continues at this constant speed.

Tram B starts at $t=20$, accelerates at $3\ \mathrm{m\,s^{-2}}$ for $4$ s, reaching speed $12\ \mathrm{m\,s^{-1}}$, then continues at this constant speed. Its graph crosses A's distance later.

Tram A's constant speed is $10$. Tram B reaches speed $10$ after $\dfrac{10}{3}$ s of acceleration, so the time from $t=0$ is $20+\dfrac{10}{3}=\dfrac{70}{3}$ s.

A travels $\dfrac12(5)(10)=25$ m in its acceleration phase, then $10\left(\dfrac{70}{3}-5\right)=\dfrac{550}{3}$ m. Total $=\dfrac{625}{3}$ m.

B is still accelerating from rest, so its distance is $\dfrac12\left(\dfrac{10}{3}\right)(10)=\dfrac{50}{3}$ m.

$d=\dfrac{625}{3}-\dfrac{50}{3}=\dfrac{575}{3}$ m.

For $t>24$, A's distance is $25+10(t-5)=10t-25$. B's distance is $\dfrac12(4)(12)+12(t-24)=12t-264$.

Set $10t-25=12t-264$. Then $2t=239$ and $t=119.5$ s.

Using A's distance, $10(119.5)-25=1170$ m.

A bearing of $150^\circ$ is $30^\circ$ east of south. Therefore the east component is $40\sin150^\circ=20$ and the north component is $40\cos150^\circ=-20\sqrt3$.

$\mathbf v_B=20\mathbf i-20\sqrt3\mathbf j\ \mathrm{km\,h^{-1}}$.

Boat A starts at $60\mathbf i$ and travels due west at $20\ \mathrm{km\,h^{-1}}$, so $\mathbf r=60\mathbf i-20t\mathbf i=(60-20t)\mathbf i$.

Boat B starts at O, so $\mathbf s=t(20\mathbf i-20\sqrt3\mathbf j)=20t\mathbf i-20\sqrt3t\mathbf j$.

$\overrightarrow{AB}=\mathbf s-\mathbf r=(20t-(60-20t))\mathbf i-20\sqrt3t\mathbf j=(40t-60)\mathbf i-20\sqrt3t\mathbf j$.

At noon the distance between the boats is $60$ km. Set the later distance equal to $60$: $(40t-60)^2+(-20\sqrt3t)^2=60^2$.

Expanding gives $2800t^2-4800t=0$, so $t(2800t-4800)=0$. The later time is $t=\dfrac{4800}{2800}=\dfrac{12}{7}$ hours.

$\dfrac{12}{7}$ hours is $1$ hour and $42.857\ldots$ minutes after noon, which rounds to $1{:}43$ pm, or $13{:}43$.