June 2025 Paper 1 Q13

13 The displacement \(\mathbf{r}\) m of a parachutist \(t\) s after opening their parachute is modelled by

\[\mathbf{r} = \begin{pmatrix} 50t \\ 280 + 5t - 280\mathrm{e}^{-0.16t} \end{pmatrix}\]

where the \(x\)-direction is horizontal and the \(y\)-direction is vertically downwards.

(a) Calculate the distance from the parachutist’s initial position that the model predicts after 10 s. [3]

(b) Find a vector expression for the velocity of the parachutist according to the model. [3]

(d) Parachutists usually land travelling approximately vertically.

Explain a factor that should be included in the model to better reflect this. [1]

Mark scheme (a)

Scheme	Marks	AO
When \(t = 10\) \(\mathbf{r} = \begin{pmatrix} 500 \\ 280 + 50 - 280\mathrm{e}^{-1.6} \end{pmatrix}\)	M1	3.4
Distance \(= \|\mathbf{r}\| = \sqrt{500^2 + 273.47^2}\)	M1	3.1b
\(= 569.9\) m	A1	1.1
	[3]

M1: Finds displacement as vector or in two directions separately
Soi
Allow use of \(\mathbf{i}\) and \(\mathbf{j}\) vectors throughout

M1: Pythagoras’ used

A1: Must be scalar answer

Mark scheme (b)

Scheme	Marks	AO
\(\mathbf{v} = \begin{pmatrix} 50 \\ 5 + 44.8\mathrm{e}^{-0.16t} \end{pmatrix}\)	M1 A1 A1	3.4 1.1 1.1
	[3]

M1: Attempt to differentiate at least one component
soi

A1: One correct component or scalar in one direction

A1: Both correct components in a vector

Mark scheme (c)

Scheme	Marks	AO
[For large values of \(t,\ \mathrm{e}^{-0.16t} \approx 0\)] So velocity tends to \(\mathbf{v} = \begin{pmatrix} 50 \\ 5 \end{pmatrix}\)	B1 B1	3.4 2.2a
	[2]

B1: Either value seen

B1: Allow for vector or both components clear which is horizontal and which vertical
Do not allow if the final answer is scalar

Mark scheme (d)

Scheme	Marks	AO
The model should take into account a resistance in the horizontal (\(x\)) direction.	B1	3.5c
	[1]

B1: Must indicate the resistance has a horizontal component.
Do not accept “air resistance” on its own.

SC1 “Factor \(e^{-kt}\) should be included” or similar