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A train traveling at a speed of 
s miles per hour applies its brakes before a buffer stop. Assuming 
d >= 0 and 
s >= 0, the distance, 
d, in yards from the train to the buffer stop once the train comes to rest is:

d=0.5(-s^(2)-1.2 s+184:}
Which of the following equivalent expressions for 
d contains the traveling speed of the train, as a constant or coefficient, for which the train rests right at the buffer stop after applying its brakes?
Choose 1 answer:
(A) 
-0.5s^(2)-0.6 s+92.3
(B) 
-0.5(s-13)(s+14.2)
(C) 
-0.5(s+0.6)^(2)+92.48
(D) 
(6.5-0.5 s)(s+14.2)

A train traveling at a speed of s s miles per hour applies its brakes before a buffer stop. Assuming d0 d \geq 0 and s0 s \geq 0 , the distance, d d , in yards from the train to the buffer stop once the train comes to rest is:\newlined=0.5(s21.2s+184 d=0.5\left(-s^{2}-1.2 s+184\right. \newlineWhich of the following equivalent expressions for d d contains the traveling speed of the train, as a constant or coefficient, for which the train rests right at the buffer stop after applying its brakes?\newlineChoose 11 answer:\newline(A) 0.5s20.6s+92.3 -0.5 s^{2}-0.6 s+92.3 \newline(B) 0.5(s13)(s+14.2) -0.5(s-13)(s+14.2) \newline(C) 0.5(s+0.6)2+92.48 -0.5(s+0.6)^{2}+92.48 \newline(D) (6.50.5s)(s+14.2) (6.5-0.5 s)(s+14.2)

Full solution

Q. A train traveling at a speed of s s miles per hour applies its brakes before a buffer stop. Assuming d0 d \geq 0 and s0 s \geq 0 , the distance, d d , in yards from the train to the buffer stop once the train comes to rest is:\newlined=0.5(s21.2s+184 d=0.5\left(-s^{2}-1.2 s+184\right. \newlineWhich of the following equivalent expressions for d d contains the traveling speed of the train, as a constant or coefficient, for which the train rests right at the buffer stop after applying its brakes?\newlineChoose 11 answer:\newline(A) 0.5s20.6s+92.3 -0.5 s^{2}-0.6 s+92.3 \newline(B) 0.5(s13)(s+14.2) -0.5(s-13)(s+14.2) \newline(C) 0.5(s+0.6)2+92.48 -0.5(s+0.6)^{2}+92.48 \newline(D) (6.50.5s)(s+14.2) (6.5-0.5 s)(s+14.2)
  1. Understand equation for distance: Understand the given equation for distance dd. The given equation is d=0.5(s21.2s+184)d = 0.5(-s^2 - 1.2s + 184). We need to find an equivalent expression that represents the scenario where the train stops right at the buffer stop.
  2. Analyze answer choices: Analyze the answer choices.\newlineWe need to find an expression that is equivalent to the given equation. This means that the expression must simplify to the same quadratic equation as the given one.
  3. Check option (A): Check option (A).\newlineOption (A) is 0.5s20.6s+92.3-0.5s^2 - 0.6s + 92.3. This is not equivalent to the given equation because the constant term is different (92.392.3 instead of 0.5×184=920.5 \times 184 = 92).
  4. Check option (B): Check option (B).\newlineOption (B) is 0.5(s13)(s+14.2)-0.5(s - 13)(s + 14.2). To check if this is equivalent, we need to expand the expression:\newline0.5(s2+14.2s13s13×14.2)-0.5(s^2 + 14.2s - 13s - 13 \times 14.2)\newline=0.5(s2+1.2s183.6)= -0.5(s^2 + 1.2s - 183.6)\newlineThis is equivalent to the given equation because when we multiply by 0.5-0.5, we get the original quadratic equation: 0.5s20.6s+91.8-0.5s^2 - 0.6s + 91.8, which is not exactly the same as the given equation due to the constant term.

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