Real-life graphs and areas under graphs

(a) Go up from 4 hours to the line and across: £6.

(b) Go across from £9 to the line and down: 6 hours.

$08\,00 + 6$ hours $= 14\,00$ (2 pm)

(a) At 2 hours the pay is £30, so $30 \div 2 = £15$ per hour.

(b) $15 \times 36 = £540$

(a) Up from 2 minutes to the curve, across: 80 °C.

(b) Across from 50 °C to the curve, down: 8 minutes.

(c) From 100 to about 73 is a drop of about 27. From about 47 to 40 is a drop of about 7. So yes, Pam is right.

(a) Go up from 3 stones to the line and across: about 19 kg.

(b) 80 kg is off the graph, so use 40 kg: 40 kg is about 6.3 stones.

$80\text{ kg} = 2 \times 6.3 = 12.6$ stones

(a) The friend's house is the first flat part at 3 km; the skate park is at 5 km. $5 - 3 = 2$ km.

(b) She arrives at 3pm, stays 2 hours: flat line at 5 km until 5pm.

She then walks home in $1\frac{1}{2}$ hours: straight line from 5 km at 5pm to 0 km at 6:30pm.

(a) Go up from 6 ounces to the line and across: about $170$ g.

(b) $1$ kg $= 1000$ g, which is off the graph. Read $200$ g $\approx 7$ ounces.

$1000 = 5 \times 200$, so $1000$ g $\approx 5 \times 7 = 35$ ounces.

(a) Read across from 10 miles: 16 km.

(b) From the graph, 25 km $\approx 15.5$ miles, so 50 km $\approx 31$ miles.

Total: $17 + 31 = 48$ miles

(a)(i) The first line goes from 12:00 to 12:30, so 30 minutes.

(a)(ii) The flat section is at 10 miles.

(b) She reaches the beach at 13:30; draw a flat line at 35 miles to 15:00, then a straight line to 0 at 16:00.

(c) 35 miles in 1 hour, so $35 \div 1 = 35$ mph.

(a) Across from 850 g to the line, then down: 30 ounces.

(b) 80 is off the graph, so use 20 ounces: 20 ounces ≈ 567 g.

$80 = 4 \times 20$, so $4 \times 567 \approx 2268$ g.

The graph shows £50 = 65 dollars.

$345 \div 65 = 5.307...$

$5.307... \times 50 \approx £265$

(a) The flat part runs from 12:45 to 13:30, which is 45 minutes.

(b) He drives 25 km home between 13:30 and 14:00, which is half an hour.

Speed $= 25 \div 0.5 = 50$ km/h

(a) At 0 miles the line starts at £10, so the fixed charge is £10.

(b) The line rises from £10 to £100 over 60 miles, which is £1.50 per mile.

20 miles more costs $20 \times 1.5 = £30$ more (e.g. 20 miles £40, 40 miles £70).

(a) The line is flat from 08 30 to 08 45: 15 minutes.

(b) At 08 50 the graph is at about 4.6 km.

(c) She travels 4 km in 20 minutes. 20 minutes $= \frac{1}{3}$ hour.

Speed $= 4 \div \frac{1}{3} = 12$ km/h

(a) 20 km in 15 minutes. There are four lots of 15 minutes in an hour, so $20 \times 4 = 80$ km/h.

(b) Stop: flat line at 20 km from 10 15 to 10 25.

In 20 minutes at 75 km/h: $75 \times \frac{20}{60} = 25$ km.

Draw a line from (10 25, 20) to (10 45, 45).

(a) 36 minutes $= 0.6$ hours, so distance $= 15 \times 0.6 = 9$ km.

Plot (9 am, 0) to (9.36 am, 9), then on to (10.45 am, 17).

(b) 15 minutes $= 0.25$ hours

Distance $= 18 \times 0.25 = 4.5$ km

(a) 4 cups of rice need 5 cups of water, and 0 need 0. Plot $(0, 0)$, $(4, 5)$ and $(8, 10)$ and join with a ruler.

(b)(i) Gradient $= \frac{\text{rise}}{\text{run}} = \frac{5}{4} = 1.25$

(b)(ii) It is the number of cups of water used for every cup of rice.

(a) The line passes through $(0, 0)$ and $(100, 14)$.

Gradient $= \frac{14}{100} = 0.14$

(b) The gradient is pounds per unit, so it is the price of one unit of electricity: 14p.

(a) From $(0, 4)$ to $(8, 16)$: gradient $= \frac{16-4}{8-0} = \frac{12}{8} = 1.5$

(b) The volume rises 1.5 litres every second, so the gradient is the rate the container fills, in litres per second.

(c) At $t = 0$ there are already 4 litres, so the intercept is the starting volume.

(a) The line goes from (0, 210) to (140, 0).

Gradient $= \frac{0 - 210}{140 - 0} = -1.5$

(b) The depth of water goes down by 1.5 cm every second: it is the rate the tank is emptying.

(a) The line starts at 27 litres when $d = 0$ and reaches 0 litres at $d = 300$.

Gradient $= \frac{0 - 27}{300 - 0} = -0.09$.

(b) The car uses 0.09 litres of petrol for each kilometre it travels (the rate of fuel consumption).

(a) Draw a tangent to the curve at $t = 5$. For example it drops about 10 m/s over 4 seconds, so gradient $\approx -10 \div 4 = -2.5$ m/s².

(b) Readings: $v(0)=25, v(2)=16, v(4)=9, v(6)=4$. Three trapezia of width 2:

$\frac12 \times 2 \times (25+16) + \frac12 \times 2 \times (16+9) + \frac12 \times 2 \times (9+4) = 41 + 25 + 13 = 79$ m

Split 0–30 s into strips of width 10. Read velocities: 0, 54, 76, 92 m/s.

Triangle: $\frac{1}{2} \times 10 \times 54 = 270$

Trapezium: $\frac{1}{2} \times 10 \times (54 + 76) = 650$

Trapezium: $\frac{1}{2} \times 10 \times (76 + 92) = 840$

Distance $\approx 270 + 650 + 840 = 1760$ m

(a) Draw a tangent touching the curve at $t = 2$. Pick two points on it, e.g. $(0, 1)$ and $(3, 3.7)$.

Gradient $\approx \frac{3.7 - 1}{3 - 0} = \frac{2.7}{3} = 0.9$

(b) Area under a speed-time graph gives the distance travelled.

(a) Read the speeds: at $t = 1, 2, 3, 4$ they are $4, 6, 7.2, 7.8$.

$\frac{1}{2}(4 + 6) + \frac{1}{2}(6 + 7.2) + \frac{1}{2}(7.2 + 7.8) = 5 + 6.6 + 7.5 = 19.1$

(b) Area under a speed–time graph is the distance travelled, so this is the distance in metres between 1 and 4 seconds.

(a) At 30 s the speed is about 9 m/s. Treat the area as a triangle: $\frac{1}{2} \times 30 \times 9 = 135$ m

(b) The curve bulges above the straight line from $(0,0)$ to $(30,9)$, so the triangle misses some area: an underestimate.

(c) Speed ÷ time gives the average acceleration from 0 to 60 s. The acceleration at 60 s is the gradient of the tangent to the curve at that point.

(a)(i) Draw a tangent at $t = 15$. Reading two points on it, e.g. it rises 7.5 over 9 s:

gradient $\approx 7.5 \div 9 = 0.83$

(ii) This is the car's acceleration in m/s² at 15 seconds.

(b) Speeds at $t = 0, 5, 10, 15, 20$: $0, 4, 12, 18, 20$.

Area $= \frac{5}{2}(0+4) + \frac{5}{2}(4+12) + \frac{5}{2}(12+18) + \frac{5}{2}(18+20) = 10 + 40 + 75 + 95 = 220$ m

(a) Distance = area: $\frac{1}{2}(30)(12) + 50(12) + \frac{1}{2}(20)(12) = 180 + 600 + 120 = 900$ m

Half distance $= 450$ m. After 30 s the train has gone 180 m, so it needs $270$ m more at 12 m/s: $270 \div 12 = 22.5$ s.

Time $= 30 + 22.5 = 52.5$ s

(b) First part: $12 \div 30 = 0.4\text{ m/s}^2$. Last part: $-12 \div 20 = -0.6\text{ m/s}^2$. The train slows down more sharply than it speeds up.