# Two automobiles are travelling on intersecting roads at 35 km/h and 25 km/h. Assuming that these cars will continue at this rate, will they collide?Two automobiles are travelling on intersecting...

Two automobiles are travelling on intersecting roads at 35 km/h and 25 km/h. Assuming that these cars will continue at this rate, will they collide?

Two automobiles are travelling on intersecting roads. The first automobile is travelling northeast at 35 km/h. The second automobile begins 7 km north of the first, and it is travelling east at 25 km/h. Assuming that these cars will continue at this rate, will they collide? Assign vector equations to each line and assume that the first automobile begins at the origin.

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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We can model the roads with the following graph:

The first road heads northeast and goes through the origin (the origin is arbitary but we are told to put it where the first car starts driving).  That is the line `y=x`

The second car starts 7km north of the first, so that is the point (0,7) on our graph.  It travels east, so we can model that by the line `y=7`

The point of intersection is (7,7)

How long will it take each car to get there?

The distance from the origin to the point (7,7) is: `7sqrt(2)=9.8994949`

Traveling at 35kph, it takes the first car `9.8994949/35=.28284271`

hours to get to the intersection point

The distance traveled by the second car is just 7km

Traveling at 25 kph, it takes the second car `7/25=.28` hours

Thus, they arrive at the intersection point .00284271 hours apart, or  .17056 minutes apart, or 10.23376 seconds apart.

The second (slower) car arrives first.  In that 10 extra seconds (.00284271 hours) it will travel .00284271*25=.07106 km, or 71 meters.  Assuming the car is not giant sized (well under 71 meters), the cars will not collide.

The vector equations we can assign to model this situation are:

The second car starts at (0,7) and travels east at 25 kph, so:

`(x,y)=(25t,7)` where t is given in hours, x and y are in km

The first car starts at (0,0) and travels NE at 35 kph, so:

`(x,y)=( ((35)/(sqrt(2))) t, ((35)/(sqrt(2))) t)`

(To see why this is so, think of an isoceles triangle with hypotenuse 35.  The legs must then have length `(35)/(sqrt(2))` .  So, while the car has traveled 35 km every hour, in each of the x and y directions it has only traveled `(35)/(sqrt(2))` km)