At time t = 0, a parachutist jumps out of an airplane that is travelling horizontally. The velocity, v m s^-1, of the parachutist at time t seconds is given by: v = (40e^{-0.2t})i + 50(e^{-0.2t} - 1)j The unit vectors i and j are horizontal and vertical respectively. Assume that the parachutist is at the origin when t = 0. Model the parachutist as a particle. (a) Find an expression for the position vector of the parachutist at time t. (b) The parachutist opens her parachute when she has travelled 100 metres horizontally. Find the vertical displacement of the parachutist from the origin when she opens her parachute. (c) Carefully, explaining the steps that you take, deduce the value of g used in the formulation of this model.

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At time t = 0, a parachutist jumps out of an airplane that is travelling horizontally - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 2

Question 15

At time t = 0, a parachutist jumps out of an airplane that is travelling horizontally. The velocity, v m s^-1, of the parachutist at time t seconds is given by: v = ... show full transcript

Worked Solution & Example Answer:At time t = 0, a parachutist jumps out of an airplane that is travelling horizontally - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 2

Step 1

Find an expression for the position vector of the parachutist at time t.

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Answer

To find the position vector, we need to integrate the velocity vector with respect to time:

$r(t) = \int v \, dt$

Given the velocity: $v = (40e^{-0.2t})i + 50(e^{-0.2t} - 1)j$

Integrating each component:

For the horizontal component:
$x(t) = \int 40e^{-0.2t} \, dt = -200e^{-0.2t} + C_x$ Setting C_x = 0 since the parachutist starts at the origin gives us: $x(t) = -200e^{-0.2t}$
For the vertical component:
$y(t) = \int 50(e^{-0.2t} - 1) \, dt = 250e^{-0.2t} - 50t + C_y$ Setting C_y = 0 gives: $y(t) = 250e^{-0.2t} - 50t$

Combining both components, the position vector is: $r(t) = (-200e^{-0.2t})i + (250e^{-0.2t} - 50t)j$

Step 2

Find the vertical displacement of the parachutist from the origin when she opens her parachute.

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Answer

To find the time when the parachutist has travelled 100 metres horizontally, we set:

$-200e^{-0.2t} = 100$

Solving for $t$ gives:

Rearranging: $e^{-0.2t} = -\frac{100}{200} = 0.5$
Taking the natural log: $-0.2t = \ln(0.5)$
Thus, $t = -\frac{\ln(0.5)}{0.2} \approx 3.4657$

Now substituting $t$ back into the vertical position equation to find the vertical displacement:

y(3.4657) = 250e^{-0.2 \cdot 3.4657} - 50(3.4657)

Calculating: $y(3.4657) \approx 250(0.5) - 173.285 \approx 125 - 173.285 = -48.3$

This indicates the parachutist has a vertical displacement of approximately 50 metres below the origin.

Step 3

Carefully, explaining the steps that you take, deduce the value of g used in the formulation of this model.

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Answer

First, we analyze the vertical component of the velocity:

$v_y = 50(e^{-0.2t} - 1)$

The equation describes the change in vertical velocity due to gravitational acceleration. The term $-1$ signifies the initial condition, where the motion starts downwards due to gravity. As $e^{-0.2t}$ approaches 0, the vertical component approaches: $v_y \to -50 \implies \text{the net acceleration is due to gravity } g$

Differentiating the function:

$a_y = \frac{dv_y}{dt} = -10e^{-0.2t}$

At $t = 0$ (just after the jump):
$a_y = -10$
This implies that the acceleration due to gravity $g$ is approximately 10 m/s², as the other forces are negligible.

At time t = 0, a parachutist jumps out of an airplane that is travelling horizontally - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 2

Worked Solution & Example Answer:At time t = 0, a parachutist jumps out of an airplane that is travelling horizontally - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 2

Find an expression for the position vector of the parachutist at time t.

Find the vertical displacement of the parachutist from the origin when she opens her parachute.

Carefully, explaining the steps that you take, deduce the value of g used in the formulation of this model.

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