A race car driving under the caution flag at 40 feet per second begins to accelerate at a constant rate after the warning flag.
The distance traveled since the warning flag in feet is characterized by 30t2 + 40t, where t is the time in seconds after the car starts accelerating again. How long will the car take to travel 150 feet?
1 answer:
Answer:
t = 1.667 s
Step-by-step explanation
The distance traveled since the warning flag in feet is characterized by
d = 30*t^2 + 40*t
Where t is the time in seconds after the car starts accelerating.
We can easily solve this question by plotting the equation using a graphing calculator or plotting tool.
We need to find the time for which the distance d = 150 ft
150 = 30*t^2 + 40*t , t > =0
We can see that this value in the graph is approximately
t = 1.667 s
We can verify
30*(1.667)^2 + 40*(1.667 ) ≈ 150
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Answer:
The value of x that gives the maximum transmission is 1/√e ≅0.607
Step-by-step explanation:
Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.
We need to equalize f' to 0
- k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
- 2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
- 2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
- ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
- 1/x = e^(1/2)
- x = 1/e^(1/2) = 1/√e ≅ 0.607
Thus, the value of x that gives the maximum transmission is 1/√e.