Question

v: velocity (mph)	20	30	40	50	60	70	80
d: stopping distance (feet)	39	75	118	172	230	305	395

The table above gives the stopping distance d of a car driven at velocity v by an unimpaired person on a paved road in good weather. The function d(v), measured in feet, is twice differentiable with d'(v)>0 and d"(v)>0.

1) Find the average rate of change of the stopping distance of the car between 50 and 60 mph. Specify units.

2) Use linear approximation to estimate the stopping distance of a car traveling at 90 mph. Based on the given information, is the actual stopping distance of the car greater than or less than this figure? Justify

3) The function d can be modeled using d=0.05v² + 0.86v + 2.5. Find the difference between the instantaneous rate of change of stopping distance between someone traveling 25 mph and someone traveling at 65 mph. Then find the difference between the instantaneous rate of change of stopping distance between someone traveling at x mph and someone traveling at y mph. Specify units.

4) The additional stopping distance d_t(v) in feet of a person who is texting while driving is modeled by d_t(v)=v. Using the model from part 3, find the ratio of stopping distance of someone texting traveling at 50 mph to someone not texting traveling at the same speed.

5) The additional stopping distance d_r(v) in feet of a person who is reading email while driving is modeled by d_r(v)=0.6v-6. Using the model from part 3, find the stopping distance of someone traveling at 50 mph and reading email and how fast this distance is changing.

Answer 1

1.)

The average rate of change of the stopping distance of the car between 50 and 60 mph is,

$d^{roc}_{avg}=\frac{\Delta d}{\Delta v}=\frac{230-172}{60-50}=5.8 \;\;feet/mph$

2.)

If we look at the last two data point,

The average rate of change of the stopping distance of the car between 70 and 80 mph is,

$d^{roc}_{avg}=\frac{\Delta d}{\Delta v}=\frac{395-305}{80-70}=9\;\;feet/mph$

Lets, assume this was the rate of stopping distance at 80 mph

Therefor,

Using linear approximation,

$d(90)-d(80) \approx d^{roc}_{avg}(75)*(90-80)\\ \Rightarrow d(90) \approx 395+9*10=485 feet$

Now, this approximate stopping distance will be less than actual stopping distance as ${d}''(v)>0$.

This means as v increases, rate of change of stopping distance also increases. Thus it will be greater than linear approximation.

3)

Instantaneous rate of change of stopping distance is ${d}'(v)$

As

${d}(v)=0.05v^2+0.86v+2.5$

differentiating

${d}'(v)=2*0.05v+0.86=0.1v+0.86$

Therefor

${d}'(65)=0.1*65+0.86=7.36feet/mph$

and

${d}'(25)=0.1*25+0.86=3.36feet/mph$

so,

${d}'(65)-{d}'(25)=4feet/mph$

simillarly,

${d}'(x)-{d}'(y)=0.1x+0.86-0.1y-0.86 =0.1(x-y)$

considering x is greater than y.

4)

$d_{text}(v)=d(v)+dt(v)=0.05v^2+0.86v+2.5+v=0.05v^2+1.86v+2.5$

Therefor,

$\frac{d_{text}(50)}{d(50)}=\frac{0.05*50^2+1.86*50+2.5}{0.05*50^2+0.86*50+2.5}=\frac{220.5}{170.5}=1.29$

5)

$d_{email}(v)=d(v)+dr(v)=0.05v^2+0.86v+2.5+0.6v-6\\ \Rightarrow d_{email}(v)=0.05v^2+1.46v-3.5$

therefor,

$d_{email}(50)=0.05*50^2+1.46*50-3.5=194.5 feet$

${d}'_{email}(v)=0.1v+1.46$

therefor,

${d}'_{email}(50)=0.1*50+1.46=6.46 feet/mph$