. 3) Any time beyond t=5.28 will not accurately reflect the car's path, because at that moment, it will have reached a speed of 0ft/s, and without an external force, the car will remain stationary past that point. 4)
(refer to attached image for the graph). Step-by-step breakdown: 1) Here, we need to determine when the vehicle's speed becomes 0, indicating a complete stop. We achieve that by deriving the position function:
. Setting the first derivative to zero gives us:
95-18t=0, from which we solve for t: -18t=-95, leading to t=5.28s. Now we will calculate the position of the vehicle after 5.28 seconds:
This results in the vehicle stopping 250.69ft after brake application, leaving approximately 50ft between the vehicle and the cow when it halts completely, allowing reliance solely on the brakes. 2) For part 2, I derive the function again to obtain:
j"(t)=-18 and then graph them (see attached image). Here, j(t) illustrates the distance post-brake application t seconds later in feet, j'(t) signifies the velocity t seconds after brakes were applied in ft/s, while j"(t) indicates the car's acceleration post-braking in
. 3) Yes, after t=5.28, the models in part (ii) won't accurately represent the vehicle's trajectory, as at that instance the car's speed would be 0ft/s and if no additional force acts on it, the vehicle won't move beyond that point. 4)
(refer to the attached image for the graph).
