Question 1190843
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Part (a)


Add the reaction distance and braking distance functions to get the overall stopping distance
D(v) = R(v) + B(v)
D(v) = 2.2v + 0.05v^2 + 0.4v - 15
D(v) = 0.05v^2 + 2.6v - 15


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Part (b)


In general, the domain is the set of all allowed inputs of the function.


More specifically, the domain is the set of real numbers on the interval *[tex \large 0 \le \text{v} \le M] where M is the max speed of the car. 


This max speed is not stated in the instructions, so we cannot describe the domain's endpoints purely in a numeric sense. You may have to contact your teacher for clarification.


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Part (c)


Plug v = 60 into the result found in part (a)
D(v) = 0.05v^2 + 2.6v - 15
D(60) = 0.05(60)^2 + 2.6(60) - 15
D(60) = 321


The stopping distance is 321 feet.


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Part (d)


The D(v) function found in part (a) is the stopping distance in feet.


To convert from feet to miles, we divide by 5280.


D(v) = stopping distance in feet
E(v) = stopping distance in miles
E(v) = ( D(v) )/5280
E(v) = (0.05v^2 + 2.6v - 15)/5280


Then if we wanted to plug in v = 60 for instance, then,
E(v) = (0.05v^2 + 2.6v - 15)/5280
E(60) = (0.05(60)^2 + 2.6(60) - 15)/5280
E(60) = 321/5280
E(60) = 0.06079545454546


The stopping distance of 321 feet is roughly equivalent to 0.0608 miles.
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