Graph gradients and areas

(a) $33$ m

1. $$(1+5)\times 2/2$$
2. $$(5+8)\times 2/2+(8+6)\times 2/2$$
3. Distance is the area under a speed-time graph. A trapezium has area $\frac{1}{2}$(a + b)h.
4. The three areas are $\frac{1}{2}$(1 + 5) $\times$ 2 = 6, $\frac{1}{2}$(5 + 8) $\times$ 2 = 13 and $\frac{1}{2}$(8 + 6) $\times$ 2 = 14.
5. The estimated distance is 6 + 13 + 14 = 33 metres.

• P1 Establishing $(1+5)\times 2/2$ or an equivalent valid method.
• P1 Establishing $(5+8)\times 2/2+(8+6)\times 2/2$ or an equivalent valid method.
• A1 Correct answer: $33$ m

(a) $6$ m/s

1. The speed is the gradient of the tangent: $\frac{26 - 2}{5 - 1} = \frac{24}{4} = 6$ m/s.

• M1 Finding the gradient of the tangent.
• A1 The correct answer, $6$ m/s.

(b) The tangent is drawn by eye, so its gradient is only approximate.

1. The tangent is drawn by eye, so the points read from it, and therefore its gradient, are approximate.

• C1 Saying the tangent is drawn by eye (or its position is approximate).

(a) $30$ m

1. Find the areas of the acceleration triangle and constant-speed rectangle.
   $$2\times 5/2+3\times 5$$
2. Add the area of the deceleration triangle.
   $$20+4\times 5/2$$
3. Therefore $30$ m.

• M1 Find the areas of the acceleration triangle and constant-speed rectangle.
• M1 Add the area of the deceleration triangle.
• A1 Correct answer: $30$ m

(b) $1.25$ m/s²

1. Divide the drop in speed by the elapsed time.
   $$5/(9-5)$$
2. Therefore $1.25$ m/s².

• M1 Divide the drop in speed by the elapsed time.
• A1 Correct answer: $1.25$ m/s²

(a) $49$ m

1. Use width 2 and half the sum of endpoint heights for each trapezium.
   $$(3+7)\times 2/2$$
2. Add all three trapezium areas.
   $$(3+2\times 7+2\times 10+12)\times 2/2$$
3. Therefore $49$ m.

• P1 Use width 2 and half the sum of endpoint heights for each trapezium.
• P1 Add all three trapezium areas.
• A1 Correct answer: $49$ m

(b) No. The trapezia join measurements with straight lines; a curved speed graph can enclose a different area between measurements.

1. No. The trapezia join measurements with straight lines; a curved speed graph can enclose a different area between measurements.

• C1 Correct conclusion with supporting reasoning: No. The trapezia join measurements with straight lines; a curved speed graph can enclose a different area between measurements.

(a) $4$ m/s

1. Use the gradient of the tangent, not a line from the origin.
   $$\frac{28-4}{7-1}$$
2. Therefore $4$ m/s.

• M1 Use the gradient of the tangent, not a line from the origin.
• A1 Correct answer: $4$ m/s

(b) $4$ m/s

1. Average speed uses the change in distance over the whole interval.
   $$\frac{18-2}{4}$$
2. Therefore $4$ m/s.

• M1 Average speed uses the change in distance over the whole interval.
• A1 Correct answer: $4$ m/s

(a) $150$ litres

1. Find the area of the triangle for the first 5 minutes.
   $$1/2\times 5\times 12$$
2. Add the area of the rectangle for the next 10 minutes.
   $$30+10\times 12$$
3. Therefore $150$ litres.

• P1 Find the area of the triangle for the first 5 minutes.
• P1 Add the area of the rectangle for the next 10 minutes.
• A1 Correct answer: $150$ litres

(b) $2.4$ litres per minute per minute

1. Find the gradient of the first section.
   $$12/5$$
2. Therefore $2.4$ litres per minute per minute.

• P1 Find the gradient of the first section.
• A1 Correct answer: $2.4$ litres per minute per minute

(c) The total volume of water that has flowed into the tank.

1. The total volume of water that has flowed into the tank.

• C1 Correct conclusion with supporting reasoning: The total volume of water that has flowed into the tank.

(a) $2.5$ m/s²

1. Divide the change in velocity by the time taken.
   $$\frac{30-10}{8}$$
2. Therefore $2.5$ m/s².

• P1 Divide the change in velocity by the time taken.
• A1 Correct answer: $2.5$ m/s²

(b) $160$ m

1. Find the area of the trapezium under the velocity-time graph.
   $$(30+10)/2\times 8$$
2. Therefore $160$ m.

• P1 Find the area of the trapezium under the velocity-time graph.
• A1 Correct answer: $160$ m

(c) $36$ km/h

1. Multiply by 3600 seconds and divide by 1000 metres.
   $$10\times 3600/1000$$
2. Therefore $36$ km/h.

• M1 Multiply by 3600 seconds and divide by 1000 metres.
• A1 Correct answer: $36$ km/h

(a) $T = 10$

1. The distance is the area under the velocity-time graph.
   $$\tfrac{1}{2} \times T \times 15 + 40 \times 15 + \tfrac{1}{2} \times 10 \times 15 = 750$$
2. $$7.5T + 600 + 75 = 750 \Rightarrow 7.5T = 75 \Rightarrow T = 10$$

• P1 Recognising that distance is the area under the velocity-time graph.
• P1 Finding the area of the constant-speed and slowing sections: 600 and 75.
• P1 Forming $7.5T + 675 = 750$.
• A1 The correct answer, $T = 10$.