QUESTION TWO (15 MARKS) (a) Consider a rock that is launched from the ground at a speed of \( 96 \mathrm{ft} / \mathrm{sec} \). Neglecting air resistance, a well known formula from physics states that the position of the rock after \( t \) seconds is given by the function \( S(t)=-16 t^{2}+96 t \). The position is measured in feet with \( S=0 \) corresponding to the ground. Find the average velocity of the rock between each pair of intervals given as i) \( t=1 \mathrm{sec} \) and \( t=3 \mathrm{sec} \), ii) \( t=1 \mathrm{sec} \) and \( t=1.00001 \mathrm{sec} \), iii) Make a conjecture regarding your observation in (i) and (ii), (1 mark) iv) By clearly showing your working evaluate \( \lim _{t \rightarrow 1} \frac{S(t)-S(1)}{t-1} \). 1 mark) (2 marks)

To find the average velocity of the rock for each specified interval, we can use the formula for average velocity, which is given by: \[ \text{Average Velocity} = \frac{S(t_2) - S(t_1)}{t_2 - t_1} \] In this case, for each pair of intervals, we will calculate \( S(t) \) first. i) For \( t=1 \) sec and \( t=3 \) sec: \[ S(1) = -16(1^2) + 96(1) = 80 \text{ ft} \] \[ S(3) = -16(3^2) + 96(3) = 192 \text{ ft} \] Average velocity between \( t=1 \) and \( t=3 \): \[ \text{Average Velocity} = \frac{S(3) - S(1)}{3 - 1} = \frac{192 - 80}{2} = \frac{112}{2} = 56 \text{ ft/sec} \] ii) For \( t=1 \) sec and \( t=1.00001 \) sec: \[ S(1.00001) = -16(1.00001^2) + 96(1.00001) \approx -16(1) + 96(1.00001) = -16 + 96.00096 = 80.00096 \text{ ft} \] Average velocity between \( t=1 \) and \( t=1.00001 \): \[ \text{Average Velocity} = \frac{S(1.00001) - S(1)}{1.00001 - 1} = \frac{80.00096 - 80}{0.00001} \approx \frac{0.00096}{0.00001} = 96 \text{ ft/sec} \] iii) From the observations in (i) and (ii) above, it can be conjectured that as the interval becomes smaller (from 2 seconds to 0.00001 seconds), the average velocity approaches the instantaneous velocity at \( t=1 \), which appears to be \( 96 \text{ ft/sec} \), the original launch speed. iv) To evaluate \( \lim _{t \rightarrow 1} \frac{S(t)-S(1)}{t-1} \), first find \( S(1) \) again: \[ S(1) = 80 \] We can re-evaluate \( S(t) \): \[ S(t) = -16t^2 + 96t \] Thus, \[ S(t) - S(1) = (-16t^2 + 96t) - 80 \] \[ = -16t^2 + 96t - 80 \] Now, compute the limit: \[ \lim_{t \rightarrow 1} \frac{-16t^2 + 96t - 80}{t-1} \] Factoring the numerator (using synthetic division or polynomial long division) gives: \[ = -16(t-1)(t+5) \div (t-1) \rightarrow -16(t+5) \] Evaluating at \( t=1 \): \[ = -16(1+5) = -16 \cdot 6 = -96 \] The limit evaluates to \( -96 \text{ ft/sec} \). The negative sign indicates the rock is moving downward towards the ground at that instant.