Higher June 2022 Paper 1 Q25
25. A hummingbird flies at a speed of \(v\) metres per second for \(t\) seconds after it has finished feeding from a flower.
The diagram shows the speed of the hummingbird for \(5 \leqslant t \leqslant 15\).
Diagram not drawn to scale
When \(t = 5\), \(v = 6\).
The average acceleration of the hummingbird for \(5 \leqslant t \leqslant 15\) is 0·5 m/s2.
Calculate the speed of the hummingbird when \(t = 15\).
You must show all your working. [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{v - 6}{15 - 5}\) | S1 | Forms a gradient in terms of \(v\) |
| \(\dfrac{v - 6}{15 - 5} = 0.5\) | M1 | FT ‘their \(\dfrac{\text{vertical diff}}{\text{horizontal diff}}\)’; implies the S1 |
| \((v =)\ 11\) (m/s) | A1 |
Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| \(v = u + at\) with \(u = 6,\ a = 0.5,\ t = 10\) | S1 | |
| \(6 + (0.5)(10)\) | M1 | Implies the S1 |
| \((v =)\ 11\) (m/s) | A1 |
Alternative method 2
| Scheme | Marks | Comments |
|---|---|---|
| Right-angled triangle drawn with horizontal 10 marked or use of \(a = \dfrac{\text{change in } v}{\text{change in } t}\) | S1 | |
| \(0.5\times10\ (+\,6)\) | M1 | Implies the S1 |
| \((v =)\ 11\) (m/s) | A1 |
Alternative method 3
| Scheme | Marks | Comments |
|---|---|---|
| \(v = 0.5t + 3.5\) | S1 | Finds the equation of the line; allow in terms of \(y\) and \(x\): \((y = v,\ x = t)\) \(y = 0.5x + c\) \(6 = 0.5(5) + c\) \(c = 3.5\) |
| \(v = 0.5(15) + 3.5\) | M1 | |
| \((v =)\ 11\) (m/s) | A1 |
| (3) |