Speed of 2 cars

Question

car A and b start at same point. A travels north 10 mph faster then b. B travels west. After 1 hour they are 108 miles apart (straight line drawn between them)

how fast is each car driving? 

So I end up with a right triangle with side a being x + 10. And side b being x. Hyp= 108 miles. 

I have tried doing this a few ways and keep getting a wrong answer. Any help/advice would be greatly appreciated. 

Answer 1

Rob, you pretty much had it.  From your

triangle use Pythagorean Theorem to solve 

for x which is the speed of car b.

Then x +10 is speed of car a:

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This creates a right triangle with hypotenuse

length of 108.

 

We need to find the leg lengths.  Using

distance = rate*time ... d = rt

 

Distance car b: d = bt

Distance car a: d = (b+10)t

Since the time is 1 hour, t = 1 giving:

Distance car b: d = b

Distance car a: d = b+10

 

Note: Above is what you had already figured out.

Now just continue from there.

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Pythagorean Theorem says:

Sum of squares of the legs of a right triangle equals

the square of the hypotenuse:

 

b² + (b+10)² = 108²

b² + b² + 20b + 100 = 11664

2b² + 20b - 11554 = 0

We can factor out a 2 since all numbers are even:

b² + 10b - 5782 = 0

 

Note that this is a quadratic so we can use

the quadratic equation: x = [-b±√(b²-4ac)]/2a

with a=1, b=10, c = -5782

 

b = [-10 ±√(10²-4(1)(-5782))]/2

b = (-10 ±√23228)/2

b = (-10 ± 152.4)/2

 

since our rate will be positive we can distegard

the negative answer giving:

 

b = (-10+152.4)/2 = 142.2/2 = 71.1 mph

 

Car a goes 71.1+10 = 81.1. mph