Question

Laura will go for a run during her lunch break if the temperature is between 6 0 ? F 60 ? F and 8 0 ? F 80 ? F. The average speed, S S, in miles per hour, at which Laura runs is dependent on the temperature, t t, in degrees Fahrenheit ( F ) ( ? F), at the start of her run and can be modeled by the function S ( t ) = 6 + 0 . 1 ( 8 0 ? t ) S(t)=6+0.1(80?t). The distance, D D, in miles, that she can run in 3 0 30 minutes given that her average speed is x x miles per hour can be modeled by the function D ( x ) = 0 . 5 x D(x)=0.5x. Find an explicit expression that models the distance that Laura runs in 3 0 30 minutes given that it is t ? F t ? F outside at the start of her run.

Answer 1

Answer:

D(x) = 3 + 0.05*(80−t)

Step-by-step explanation:

We want to find a relation between D, distance, and t, temperature.

Distance follows the equation:

D(x) = 0.5*x

where x is average speed.

Average speed is modeled by:

S(t) = 6 + 0.1*(80−t)

Replacing x = S(t):

D(x) = 0.5*[6 + 0.1*(80−t)]

D(x) = 0.5*6 + 0.5*0.1*(80−t)

D(x) = 3 + 0.05*(80−t)

Answer 2

Answer:

$D(t)=3 \frac{miles}{hour} +0.05*\frac{miles}{hour*F}(80F-t)$

Step-by-step explanation:

The average speed in function of the temperature:

$S(t)=6 \frac{miles}{hour} +0.1*\frac{miles}{hour*F}(80F-t)$

The distance in function of the average speed:

$D(x)=0.5*x$

x and S are the same

$D(t)=0.5*[6 \frac{miles}{hour} +0.1*\frac{miles}{hour*F}(80F-t)]$

$D(t)=3 \frac{miles}{hour} +0.05*\frac{miles}{hour*F}(80F-t)$