| Date | May 2011 | Marks available | 3 | Reference code | 11M.2.hl.TZ2.3 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | How far | Question number | 3 | Adapted from | N/A |
Question
A skydiver jumps from a stationary balloon at a height of 2000 m above the ground.
Her velocity, \(v{\text{ m}}{{\text{s}}^{ - 1}}\) , t seconds after jumping, is given by \(v = 50(1 - {{\text{e}}^{ - 0.2t}})\) .
Find her acceleration 10 seconds after jumping.
How far above the ground is she 10 seconds after jumping?
Markscheme
\(a = 10{{\text{e}}^{ - 0.2t}}\) (M1)(A1)
at \(t = 10\) , \(a = 1.35{\text{ }}({\text{m}}{{\text{s}}^{ - 2}})\,\,\,\,\,{\text{(accept }}10{e^{ - 2}})\) A1
[3 marks]
METHOD 1
\(d = \int_0^{10} {50(1 - {{\text{e}}^{ - 0.2t}}){\text{d}}t} \) (M1)
\( = 283.83…\) A1
so distance above ground \( = 1720{\text{ (m) (3 sf) }}\left( {{\text{accept 1716 (m)}}} \right)\) A1
METHOD 2
\({\text{s}} = \int {50(1 - {{\text{e}}^{ - 0.2t}}){\text{d}}t = 50t + 250{{\text{e}}^{ - 0.2t}}( + c)} \) M1
Taking s = 0 when t = 0 gives c = −250 M1
So when t = 10, s = 283.3...
so distance above ground \( = 1720{\text{ (m) (3 sf) }}\left( {{\text{accept 1716 (m)}}} \right)\) A1
[3 marks]
Examiners report
Part (a) was generally correctly answered. A few candidates suffered the Arithmetic Penalty for giving their answer to more than 3sf. A smaller number were unable to differentiate the exponential function correctly. Part (b) was less well answered, many candidates not thinking clearly about the position and direction associated with the initial conditions.
Part (a) was generally correctly answered. A few candidates suffered the Arithmetic Penalty for giving their answer to more than 3sf. A smaller number were unable to differentiate the exponential function correctly. Part (b) was less well answered, many candidates not thinking clearly about the position and direction associated with the initial conditions.
