Date | None Specimen | Marks available | 7 | Reference code | SPNone.2.hl.TZ0.9 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Calculate | Question number | 9 | Adapted from | N/A |
Question
A ladder of length 10 m on horizontal ground rests against a vertical wall. The bottom of the ladder is moved away from the wall at a constant speed of \(0.5{\text{ m}}{{\text{s}}^{ - 1}}\). Calculate the speed of descent of the top of the ladder when the bottom of the ladder is 4 m away from the wall.
Markscheme
let x, y (m) denote respectively the distance of the bottom of the ladder from the wall and the distance of the top of the ladder from the ground
then,
\({x^2} + {y^2} = 100\) M1A1
\(2x\frac{{{\text{d}}x}}{{{\text{d}}t}} + 2y\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0\) M1A1
when \(x = 4,{\text{ }}y = \sqrt {84} \) and \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.5\) A1
substituting, \(2 \times 4 \times 0.5 + 2\sqrt {84} \frac{{{\text{d}}y}}{{{\text{d}}t}} = 0\) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}t}} = - 0.218{\text{ m}}{{\text{s}}^{ - 1}}\) A1
(speed of descent is \(0.218{\text{ m}}{{\text{s}}^{ - 1}}\))
[7 marks]