Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.(a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 14 mi away?_ min(b) How far must the faster car travel before it has a 15-min lead on the slower car?_ mi

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JeffriI357794 JeffriI357794 · Answer 1 · 2022-10-25T16:06:14+02:00

We will have the following:

a) We first determine the time it takes to travel the distance to both vehicles:

*

[tex]t_1=\frac{14mi\ast1h}{55}\Rightarrow t_1=\frac{14}{55}h[/tex]

*

[tex]t_2=\frac{14mi\ast1h}{60mi}\Rightarrow t_2=\frac{11}{12}h[/tex]

So, we determine now the difference in time:

[tex]\frac{11}{12}h-\frac{14}{55}h=\frac{437}{660}h\approx0.66h[/tex]

So, the fastest car will arrive approximately 0.66 hours sooner.

b) We determmine the distance it must travel the fastest car to have a 15 minute lead on the other one as follows:

First, we determine the time difference required:

[tex]t=\frac{15min\ast1h}{60min}\Rightarrow t=0.25h[/tex]

Then, since both vehicles will move relative to each other, we will have that:

[tex]d_{c1}=(60mi/h)(0.25h)\Rightarrow d_{c1}=15mi[/tex]

So, the fastest car must be 15 miles ahead of the other car in order to have a 15 minute lead with respect to the second car.