A scientist is researching the effects of caffeine. She models the mass of caffeine in the body using $$m = m_0 e^{-kt}$$ where $m_0$ milligrams is the initial mass of caffeine in the body and $m$ milligrams is the mass of caffeine in the body after t hours. On average, it takes 5.7 hours for the mass of caffeine in the body to halve. One cup of strong coffee contains 200 mg of caffeine. 10 (a) The scientist drinks two strong cups of coffee at 8 am. Use the model to estimate the mass of caffeine in the scientist's body at midday. 10 (b) The scientist wants the mass of caffeine in her body to stay below 480 mg. Use the model to find the earliest time that she could drink another cup of strong coffee. Give your answer to the nearest minute. 10 (c) State a reason why the mass of caffeine remaining in the scientist's body predicted by the model may not be accurate.

Photo AI

A scientist is researching the effects of caffeine - AQA - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

Question 10

A scientist is researching the effects of caffeine. She models the mass of caffeine in the body using $$m = m_0 e^{-kt}$$ where $m_0$ milligrams is the initial ma... show full transcript

Worked Solution & Example Answer:A scientist is researching the effects of caffeine - AQA - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

Step 1

10 (a) Use the model to estimate the mass of caffeine at midday.

96%

114 rated

Answer

To find the mass of caffeine at midday, we first determine the initial mass of caffeine after consuming 2 cups of coffee:

$m_0 = 2 imes 200 = 400 ext{ mg}$

Next, we need to calculate the constant $k$ . Since it takes 5.7 hours for the mass to halve, we can state that:

$\frac{m_0}{2} = m_0 e^{-k \times 5.7}$

Substituting $m_0 = 400$ mg gives:
$200 = 400 e^{-k \times 5.7}$
This simplifies to: $e^{-k \times 5.7} = 0.5$ Taking the natural logarithm: $-k \times 5.7 = \ln(0.5)$
Thus,
$k = -\frac{\ln(0.5)}{5.7}$

Now, substituting the time from 8 am to midday (4 hours) into the model: $m = 400 e^{-k \times 4}$
By substituting the value of $k$ , we can compute $m$ :

$m \approx 400 e^{\frac{4 \ln(0.5)}{5.7}}$
Calculating this yields:

$m \approx 245.93 \text{ mg}$

Step 2

10 (b) Find the earliest time to drink another cup.

99%

104 rated

Answer

To determine when the scientist's caffeine level exceeds 480 mg, we set up the inequality based on the model:

$400 e^{-kt} < 480$

Solving the inequality, we first divide by 400: $e^{-kt} < \frac{480}{400}$

This simplifies to: $e^{-kt} < 1.2$

Taking the natural logarithm of both sides: $-kt < \ln(1.2)$

Substituting the value of $k$ : $t > -\frac{\ln(1.2)}{k}$
Using our calculated value for $k$ : $t > -5.7 \frac{\ln(1.2)}{-\ln(0.5)}$ Using $t \approx 2.93 \text{ hours}$ , the earliest time to drink another cup is: $2.93 \text{ hours after } 8 ext{ am} \approx 10:56 ext{ am}$ .

Step 3

10 (c) Reason for model accuracy issues.

96%

101 rated

Answer

The mass of caffeine predicted by the model may not be accurate because:

Different people eliminate caffeine at different rates depending on individual metabolism.
The model is based on an average person, which may not accurately represent the scientist's own caffeine clearance.
Factors such as hydration levels, age, and health can significantly affect caffeine metabolism.

A scientist is researching the effects of caffeine - AQA - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

Worked Solution & Example Answer:A scientist is researching the effects of caffeine - AQA - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

10 (a) Use the model to estimate the mass of caffeine at midday.

10 (b) Find the earliest time to drink another cup.

10 (c) Reason for model accuracy issues.

Join the A-Level students using SimpleStudy...