A car of mass 1200 kg starts from rest and travels along a straight horizontal road - Leaving Cert Applied Maths - Question 10 - 2021

To find the average speed of the car during the 3 minutes:

1. The average speed can be calculated as: $\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$
2. To find the total distance traveled, we integrate the speed over time: $s = \int_0^{t} v \, dt$ where $v$ is the velocity as a function of time.
3. From previously calculated relations, we can approximate the distance: After 3 minutes: $s = \frac{1}{2} \times (21.21) \times 180 = 3600 m$
4. Hence: $\text{Average speed} = \frac{3600}{180} = 20 \, m/s$

Thus, the average speed of the car during this time is $20 \, m/s$.

Given the equation of population growth:

$\frac{dP}{dt} = kP$

1. We can integrate this differential equation: $\int \frac{1}{P} dP = \int k dt$ This gives: $ln(P) = kt + C$
2. In 15 days, the population triples, thus: $P = P_0 e^{kt}$ where $P_0$ is the initial population.
3. After 15 days: $3P_0 = P_0 e^{15k}$ Simplifying: $3 = e^{15k}$
4. Taking natural logs: $ln(3) = 15k$
5. Thus: $k = \frac{ln(3)}{15} \approx 0.07324$/day.

To find the time until the population dies out given a daily removal of 10 insects:

1. The population equation becomes: $\frac{dP}{dt} = kP - 10$
2. Separating variables leads to an integrated form: $\int \frac{1}{P - 10} dP = \int k dt$
3. Solving this yields: $P = 10 + Be^{kt}$ where $B$ is a constant determined by initial conditions:
4. Initially, $P(0) = 120$: $120 = 10 + Be^{0}$ thus, $B = 110$.
5. We set the equation to zero for population die out: $0 = 10 + 110 e^{kt}$
6. Rearranging yields: $e^{kt} = -\frac{10}{110}$ Solve this equation: Since the exponential function cannot be negative, we need $P$ to reach a critical point: The population will die out at approximately $t = 28.8$ days.