Theresa bought a house on 2 January 1970 for £8000 - AQA - A-Level Maths Mechanics - Question 8 - 2019 - Paper 2

From the data provided, we need to analyze the relationship between $\log_{10} V$ and $t$:

1. From the table, we note the following data points:

   • When $t = 0$, $\log_{10} V = 3.90$
   • When $t = 10$, $\log_{10} V = 4.28$
   • When $t = 20$, $\log_{10} V = 4.66$
   • When $t = 30$, $\log_{10} V = 4.91$
   • When $t = 40$, $\log_{10} V = 5.31$

2. Plotting these points on a graph will allow us to visualize the linear relationship.

3. The slope of the line of best fit can be calculated using two points from the data. Let’s use (0, 3.90) and (40, 5.31):

   The slope $m$ is given by:

   $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5.31 - 3.90}{40 - 0} = \frac{1.41}{40} = 0.03525$

4. This shows that for every increase of 1 in $t$, $\log_{10} V$ increases by approximately $0.03525$.

5. Furthermore, we can estimate $p$ and $q$ from the line, noting that:

   • The intercept gives us a direct correlation to $\log_{10} p$.

   This concludes the analysis of how $\log_{10} V$ varies with $t$.