Rates of change and areas under graphs – Year 11 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Speed \(=\dfrac{60}{2}=\) 30 km/h.
   2. The graph is horizontal, so he was stationary (resting) for half an hour.
   3. He travels 60 km in \(4.5-2.5=2\) hours: 30 km/h.
   4. Total distance \(=60+60=120\) km, total time \(=4.5\) h. Average speed \(=\dfrac{120}{4.5}=\) 26.7 km/h.

2. Answer to Question 2

   1. \(72\times1000\div3600=\) 20 m/s. Multiply by 1000 to change kilometres into metres and divide by 3600 to change hours into seconds. Dividing by 3.6 does both at once.
   2. \(15\times3600\div1000=\) 54 km/h. Do the reverse: multiply by 3.6. So 15 m/s is 54 km/h.
   3. Acceleration \(=\dfrac{24-0}{8}=\) 3 m/s². Acceleration is the change in velocity divided by the time taken.
   4. Metres per second per second, m/s². The velocity changes by a number of metres per second in each second.

3. Answer to Question 3

   1. \(\dfrac{20-0}{10}=\) 2 m/s².
   2. \(20\times(30-10)=\) 400 m.
   3. Triangle \(\tfrac12\times10\times20=100\); rectangle 400; triangle \(\tfrac12\times10\times20=100\). Total \(=\) 600 m. (Check with a trapezium: parallel sides 40 s and 20 s, height 20: \(\tfrac12(40+20)\times20=600\).)
   4. \(\dfrac{600}{40}=\) 15 m/s.

4. Answer to Question 4

   1. Gradient \(=\dfrac{150}5=\) 30 litres per minute. The gradient is the change in volume divided by the change in time.
   2. The gradient is the rate of flow of the water.
   3. \(450\div30=\) 15 minutes. Divide the volume by the rate of flow.
   4. The line is less steep with gradient 20. A smaller rate gives a smaller gradient, so the line is less steep.

5. Answer to Question 5

   1. \(12\div60=\) 0.2 litres/s. There are 60 seconds in a minute, so divide by 60.
   2. \(12\times60=\) 720 litres/h. There are 60 minutes in an hour, so multiply by 60.
   3. \(30\div12=\) 2.5 minutes. Time \(=\) amount \(\div\) rate.
   4. The hose: \(0.5\times60=30\) litres per minute, which is faster than 12. Change both rates into the same units before comparing.

6. Answer to Question 6

   1. Area \(=\tfrac12\times2\times(0+6)+\tfrac12\times2\times(6+10)+\tfrac12\times2\times(10+12)=6+16+22=\) 44 m.
   2. The curve bends upwards less and less, so the straight tops of the trapezia lie below the curve: the estimate is an underestimate.
   3. \(\dfrac{44}6=\) 7.3 m/s. Average speed \(=\) total distance \(\div\) total time.

7. Answer to Question 7

   1. First: \(\dfrac{45}{1.5}=\) 30 km/h. Second: \(\dfrac{90-45}{3.5-2}=\dfrac{45}{1.5}=\) 30 km/h. Speed is the gradient of each sloping part: distance divided by time.
   2. \(2-1.5=\) 0.5 hours. The horizontal part shows no change in distance.
   3. \(\dfrac{90}{3.5}=\) 25.7 km/h. Divide the total distance by the total time, including the rest.
   4. Running time \(=1.5+1.5=3\) h. \(\dfrac{90}3=\) 30 km/h. Leave out the half-hour rest from the time.

8. Answer to Question 8

   1. First: \(\dfrac{15}{0.5}=\) 30 km/h. Last: \(\dfrac{60-15}{1}=\) 45 km/h.
   2. The bus is stopped for half an hour (a break).
   3. \(\dfrac{60}2=\) 30 km/h. Divide the total distance by the total time of 2 hours, including the stop.
   4. It took 2 hours, so it is 30 minutes late.

9. Answer to Question 9

   1. \(100\div3.6=\) 27.78 m/s. Divide by 3.6 to change km/h into m/s.
   2. \(\dfrac{27.78}8=\) 3.47 m/s². Acceleration is the change in speed divided by the time.
   3. \(3.47\times12=41.7\) m/s \(=41.7\times3.6=\) 150 km/h. Multiply the acceleration by the time, then multiply by 3.6 to change back to km/h.
   4. Distance \(=\tfrac12\times8\times27.78=\) 111 m. The distance is the area under the velocity-time graph, which is a triangle here.

10. Answer to Question 10

    1. \(2\times3=\) 6 m/s. Starting from rest, the speed increases by 2 m/s each second for 3 seconds.
    2. Corner points: \((0,0)\), \((3,6)\), \((8,6)\), \((12,0)\).
    3. Acceleration phase \(\tfrac12\times3\times6=9\) m. Constant: \(5\times6=30\) m. Deceleration phase \(\tfrac12\times4\times6=12\) m. Total \(=9+30+12=\) 51 m. The distance is the area under the velocity-time graph: two triangles and a rectangle.
    4. \(\dfrac{6}{4}=\) 1.5 m/s² (deceleration). The speed falls from 6 m/s to 0 in 4 seconds.

11. Answer to Question 11

    1. \(6\times3\times1.2=21.6\) m³ \(=\) 21 600 litres. The pool is a cuboid, so multiply the three dimensions. Then multiply by 1000 to change m³ into litres.
    2. \(21\,600\div25=864\) minutes \(=14.4\) hours \(=\) 14 hours 24 minutes. Divide the volume by the rate, then change minutes into hours: \(0.4\times60=24\) minutes.
    3. Together: \(75\) litres per minute. \(21\,600\div75=288\) minutes \(=\) 4 hours 48 minutes. Two pumps working together add their rates: \(25+50=75\) litres per minute.

12. Answer to Question 12

    1. \(24\div3.6=\) 6.67 m/s. Divide by 3.6 to change km/h into m/s.
    2. \(500\div6.67=\) 75 seconds. Time \(=\) distance \(\div\) speed, using metres and m/s.
    3. \(24\times\dfrac{45}{60}=\) 18 km. 45 minutes is \(\tfrac34\) of an hour.
    4. Runner: \(5\times3.6=18\) km/h. The cyclist is faster by \(24-18=\) 6 km/h. Change the runner's speed into km/h by multiplying by 3.6.

13. Answer to Question 13

    1. \(\dfrac{56000-50000}{2020-2010}=\) 600 people per year. The gradient is the change in population divided by the change in years.
    2. \(56\,000+10\times600=\) 62 000. Add 10 years of growth at 600 people per year.
    3. No, the gradient shows 600 per year. The rate of 1000 per year is bigger than the gradient.
    4. \(60\,000-56\,000=4\,000\), and \(4000\div600=6.7\), so in about 2027. Divide the extra people needed by the yearly increase, then add to 2020.

14. Answer to Question 14

    1. \(\dfrac{50-80}{6}=\) \(-5\)°C per minute.
    2. The negative sign shows the temperature is falling.
    3. From 80°C, falling 5 each minute: \(60\div5=\) 12 minutes.
    4. It would be a curve that gets less steep over time (a decreasing gradient).

15. Answer to Question 15

    1. \(\dfrac{12}4=\) 3 m/s². Acceleration is the change in velocity divided by the time.
    2. \(\dfrac{12}3=\) 4 m/s² (deceleration). The velocity falls from 12 m/s to 0 in 3 seconds.
    3. \(\tfrac12\times4\times12=24\); \(6\times12=72\); \(\tfrac12\times3\times12=18\). Total \(=\) 114 m. The total distance is the area under the graph: two triangles and a rectangle.
    4. \(\dfrac{114}{13}=\) 8.77 m/s. The total time is \(4+6+3=13\) seconds.

16. Answer to Question 16

    1. A: \(s=50t\). B: \(s=75(t-0.5)\) for \(t\ge0.5\).
    2. \(50t=75t-37.5\), so \(25t=37.5\) and \(t=1.5\) hours after A left.
    3. \(50\times1.5=\) 75 km.
    4. B's line starts at \((0.5,0)\) and is steeper than A's line from the origin; the lines cross at \((1.5,75)\).

17. Answer to Question 17

    1. Gradient of the tangent \(=\dfrac{11-3}{6-2}=\) 2 m/s².
    2. Area \(=\tfrac12\times2\times[(0+5)+(5+8)+(8+10)+(10+11)]=5+13+18+21=\) 57 m.
    3. The graph is concave (curves downward), so the chords lie below the curve: the estimate is an underestimate.

18. Answer to Question 18

    1. \(15-3=\) 12 litres per minute.
    2. \(400\div12=\) 33.3 minutes.
    3. \(3\times33.3=\) 100 litres.
    4. After 10 minutes: \(10\times12=120\) L. Remaining \(280\) L at 15 L/min: \(280\div15=18.7\) min. Total 28.7 minutes.

19. Answer to Question 19

    1. \(100\div4.2=\) 23.8 m/s. Average speed \(=\) distance \(\div\) time.
    2. \(23.8\times3.6=\) 85.7 km/h. Multiply by 3.6 to change m/s into km/h.
    3. The cheetah is slightly faster than 85 km/h. 85.7 km/h is a little more than 85 km/h.
    4. \(23.8\times20=\) 476 m. Distance \(=\) speed \(\times\) time.

20. Answer to Question 20

    1. \(v=0,12,16,12,0\). Substitute each value of \(t\) into \(t(8-t)\): for example \(2\times6=12\).
    2. Area \(=\tfrac12\times2\times[(0+12)+(12+16)+(16+12)+(12+0)]=12+28+28+12=\) 80 m. Each trapezium has width 2, so its area is \(\tfrac12\times2\times\) the sum of the parallel sides.
    3. True \(=85.33\). Error \(=\dfrac{85.33-80}{85.33}=\) 6.25% (underestimate). Percentage error \(=\dfrac{\text{error}}{\text{true value}}\times100\).
    4. Use more, narrower strips. Narrower strips follow the curve more closely, so the error is smaller.