Higher November 2018 Paper 2 Q28
28 Izzy runs an 80-metre race in 14 seconds.
During the first 6 seconds her speed increases at a constant rate.
During the last 8 seconds her speed increases at a different constant rate.
Her speed at 14 seconds is 2 m/s more than her speed at 6 seconds.
Here is a sketch of her speed-time graph.
Not drawn accurately
(a) Work out her acceleration during the last 8 seconds.
State the units of your answer. [2 marks]
(b) When Izzy finishes the 80-metre race, her speed is \(v\) m/s
Work out the value of \(v\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 0.25 or \(\dfrac{1}{4}\) or \(\dfrac{2}{8}\) | B1 | |
| m/s2 or ms−2 or m/s/s or \(\dfrac{\text{m}}{\text{s}^2}\) | B1 | oe eg metres per second per second SC2 acceleration and unit not in m/s2 eg 25 cm/s2 or 3240 km/h2 |
Additional guidance
| \(\dfrac{2}{14 - 6}\) with no further simplification | (1st) B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{1}{2} \times 6 \times (v - 2)\) or \(\dfrac{1}{2} \times (14 - 6) \times (v + v - 2)\) or \((14 - 6) \times (v - 2)\) or \(\dfrac{1}{2} \times (14 - 6) \times 2\) or 8 | M1 | oe partial area any letter |
| \(\dfrac{1}{2} \times 6 \times (v - 2)\) \(+\; \dfrac{1}{2} \times (14 - 6) \times (v + v - 2)\) or \(3(v - 2) + 8(v - 2) + 8\) or \(11v - 14\) | M1dep | oe full area in one variable eg \(14 \times v - \dfrac{1}{2} \times 6 \times (v - 2)\) \(-\; \dfrac{1}{2} \times 2 \times (6 + 14)\) implies M2 |
| \(\dfrac{1}{2} \times 6 \times (v - 2)\) \(+\; \dfrac{1}{2} \times (14 - 6) \times (v + v - 2) = 80\) or \(94 \div 11\) | A1 | oe full area in one variable equated to 80 |
| 8.5(4…) or 8.55 or \(\dfrac{94}{11}\) or \(8\dfrac{6}{11}\) | A1 | |
| Alternative method 2 | ||
| \(\dfrac{1}{2} \times 6 \times x\) or \(\dfrac{1}{2} \times (14 - 6) \times (x + x + 2)\) or \((14 - 6) \times x\) or \(\dfrac{1}{2} \times (14 - 6) \times 2\) or 8 | M1 | oe partial area \(x\) is the speed at 6 seconds any letter |
| \(\dfrac{1}{2} \times 6 \times x\) \(+\; \dfrac{1}{2} \times (14 - 6) \times (x + x + 2)\) or \(3x + 8x + 8\) or \(11x + 8\) | M1dep | oe full area in one variable eg \(14 \times (x + 2) - \dfrac{1}{2} \times 6 \times x\) \(-\; \dfrac{1}{2} \times 2 \times (6 + 14)\) implies M2 |
| \(\dfrac{1}{2} \times 6 \times x\) \(+\; \dfrac{1}{2} \times (14 - 6) \times (x + x + 2) = 80\) or \(72 \div 11\) or 6.5(4…) or 6.55 or \(\dfrac{72}{11}\) or \(6\dfrac{6}{11}\) | A1 | oe full area in one variable equated to 80 |
| 8.5(4…) or 8.55 or \(\dfrac{94}{11}\) or \(8\dfrac{6}{11}\) | A1 | |
Additional guidance
First M1 Do not allow 8 from 14 − 6
Ignore units throughout