Gradients and Areas Under Graphs

Consider the quantity obtained by dividing a change in velocity by a change in time.

1. The gradient is change in velocity divided by time.
2. It represents the trolley’s acceleration.

Calculate change in velocity divided by time taken.

1. The acceleration is the gradient of a velocity-time graph: change in velocity \div time taken.
2. The velocity increases from 4 m/s to 16 m/s, a change of 16-4=12 m/s, in 6 seconds.
3. The acceleration is 12\div6=2 m/s².

Distance is represented by the area under a speed-time graph.

1. The area is a rectangle of width 6 seconds and height 8 m/s.
2. The distance is 6\times8=48 m.

The region under the graph is a triangle.

1. The triangle has base 6 seconds and height 10 m/s.
2. The distance is \dfrac12\times6\times10=30 m.

Use the area of the trapezium below the line.

1. The parallel sides of the trapezium are 4 and 16, and its width is 10.
2. The distance is \dfrac12(4+16)\times10=100 m.

Split the area into two triangles and a rectangle.

1. The distance travelled is the area under the speed-time graph. Split it into a triangle, a rectangle and a triangle.
2. The coloured areas separate the quantities used in the calculation.

3. The first triangle has area \dfrac12\times4\times12=24.
4. The rectangle has area (10-4)\times12=72.
5. The final triangle has area \dfrac12\times(14-10)\times12=24.
6. The total distance is 24+72+24=120 m.

A fall in velocity gives a negative gradient.

1. The change in velocity is 2-22=-20 m/s over 10 seconds.
2. The acceleration is -20\div10=-2 m/s².

Average acceleration uses the change between the two endpoint velocities.

1. The velocity changes from 2 m/s to 18 m/s.
2. The average acceleration is (18-2)\div8=2 m/s².

Use the gradient of the tangent, keeping its sign.

1. The acceleration at an instant is estimated by the gradient of the tangent at that instant.
2. The tangent passes through (0,12) and (6,0).
3. Gradient =\dfrac{\text{change in velocity}}{\text{change in time}}=\dfrac{0-12}{6-0}=-2.
4. Use the tangent’s fall in velocity and the matching time interval.

5. The acceleration is about -2\mathrm{\,m/s^2}. It is negative because the object is slowing down.

Multiply the units of the flow rate by the units of time. For the volume, split the area into two trapezia.

Part (a)

1. Flow rate (litres per second) multiplied by time (seconds) gives litres, a volume.
2. So the area represents the total volume of water supplied.

Part (b)

1. The area under the graph splits into two trapezia, each 4 seconds wide.
2. The coloured areas separate the quantities used in the calculation.

3. From 0 to 4 seconds: \dfrac12(2+6)\times4=16 litres.
4. From 4 to 8 seconds: \dfrac12(6+2)\times4=16 litres.
5. The volume supplied is 16+16=32 litres.

Read the three heights and use a width of half the total time.

1. Each trapezium has width 4\div2=2 seconds. The speeds at t=0,2,4 are 4,6,12 m/s.

2. The first area is \dfrac12(4+6)\times2=10.
3. The second area is \dfrac12(6+12)\times2=18.
4. The estimated distance is 10+18=28 m.

Add the areas of strips with width two seconds.

1. The flow rates at t=0,2,4,6 are 12,11,8,3 litres/s.
2. The coloured areas separate the quantities used in the calculation.

3. The three areas are \dfrac12(12+11)\times2=23, \dfrac12(11+8)\times2=19 and \dfrac12(8+3)\times2=11.
4. The estimated volume is 23+19+11=53 litres.

Compare the tops of the trapezia with the curve between the endpoints.

1. Between the marked endpoints, both dashed chords lie below the curve.
2. The trapezia omit some area, so they underestimate the distance travelled.

Find the total area first, then divide by the total time.

1. The areas are two triangles of \dfrac12\times5\times10=25 m each and a rectangle of 10\times10=100 m.
2. The coloured areas separate the quantities used in the calculation.

3. The total distance is 25+100+25=150 m.
4. The average speed is 150\div20=7.5 m/s.

The time unit used in each area must match the rate’s time unit.

1. The area under a rate-time graph gives the volume. The rate is in litres per minute, so the times must be in minutes: 30 seconds =0.5 minutes.
2. The coloured areas separate the quantities used in the calculation.

3. Split the area into two trapezia, each 0.5 minutes wide, with parallel sides 60 and 180 litres/min.
4. Each trapezium has area \dfrac12(60+180)\times0.5=60 litres.
5. The total volume is 60+60=120 litres.

Convert each time width to hours before using the areas.

1. Distance is the area under the speed-time graph. The speed is in km/h, so the times must be in hours: 6 minutes =6\div60=0.1 hours.
2. The coloured areas separate the quantities used in the calculation.

3. First 6 minutes: a trapezium with parallel sides 20 and 60, so the distance is \dfrac12(20+60)\times0.1=4 km.
4. Last 6 minutes: a rectangle, so the distance is 60\times0.1=6 km.
5. The total distance is 4+6=10 km.

Express the areas of the two triangles and the rectangle in terms of the unknown speed.

1. Each triangle has area \dfrac12\times5\times V=2.5V.
2. The coloured areas separate the quantities used in the calculation.

3. The rectangle has area (15-5)V=10V.
4. The total distance gives 2.5V+10V+2.5V=225.
5. Therefore 15V=225 and V=15 m/s.

Find the total area under the graph first, then work out when the area reaches half of it.

1. Distance travelled is the area under the speed-time graph.
2. Total area: triangle \dfrac12\times4\times10=20 m, rectangle (14-4)\times10=100 m and triangle \dfrac12\times(20-14)\times10=30 m, so 150 m in all.
3. Half the distance is 150\div2=75 m.
4. By 4 seconds the tram has travelled 20 m, so it still needs 75-20=55 m. It is then moving at a steady 10 m/s.
5. The coloured areas separate the quantities used in the calculation.

6. That takes 55\div10=5.5 seconds, so half the distance is reached at 4+5.5=9.5 seconds.
7. It is not at 10 seconds, halfway through the time, because the tram takes longer to slow down than to speed up.

Compare the magnitudes of the two tangent gradients.

1. For A, the temperature falls 50-20=30 degrees over 5 minutes, giving a cooling rate of 6 degrees per minute.
2. Tangent A: use its temperature fall over the five-minute interval.

3. For B, the fall is 30-20=10 degrees over 7-2=5 minutes, giving 2 degrees per minute.
4. Tangent B: use its temperature fall over the five-minute interval.

5. The liquid is cooling about 6\div2=3 times as fast at t=2.

Find the estimated volume from pipe P, then share the rest of the 84 litres over the 8 seconds.

1. The area under a rate-time graph is the volume. Four trapezia over 8 seconds are each 2 seconds wide.
2. Pipe P's rates at t=0,2,4,6,8 are 6,9,10,9,6 litres/s.
3. The areas are \dfrac12(6+9)\times2=15, \dfrac12(9+10)\times2=19, \dfrac12(10+9)\times2=19 and \dfrac12(9+6)\times2=15.
4. The coloured areas separate the quantities used in the calculation.

5. Pipe P supplies about 15+19+19+15=68 litres.
6. Pipe Q must supply about 84-68=16 litres in 8 seconds, so its rate is 16\div8=2 litres/s.