You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

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6 days ago

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You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

your lane. The cow is fearless, and staring you down. (Or perhaps she does not understand the gravity of the situation.) Regardless, if you simply hit your brakes, after t seconds, the car will be j(t) = 95t - 9t^2 feet from the point where the brakes were first applied.
1) Must you steer to avoid the cow, or can you rely solely on your brakes? Explain.
2) Graph J(t), j'(t), and j"(t) and Interpret their meanings in context.
3) Is there a time t after which the graphs in part (ii) probably do not accurately model the path of the car?
4) Write a piece-wise function using j(t) that would more accurately model the path of the car on the interval [0, 10], assuming the car did not move after it stopped. Sketch this graph and its derivative.

Mathematics

1 answer:

lawyer [12.2K] · Answer 1 · 2026-04-04T04:15:55+03:00

1) You can depend solely on your braking system, as the vehicle will only travel 250ft from when you apply the brakes until it completely stops, leaving it 50ft away from the cow. 2) Refer to the attached image. j(t) represents the distance from the brake application point after t seconds in feet. j'(t) indicates the car's speed t seconds after braking, expressed in ft/s. j"(t) shows the car's acceleration t seconds post-braking in $ft/s^{2}$

. 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) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(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:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

. 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:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

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:

$j(t)=95t-9t^{2}$

$j'(t)=95-18t$

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 $ft/s^{2}$

. 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) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$ (refer to the attached image for the graph).