Two cars are starting from positions that are 20 miles apart...

Question

Two cars are starting from positions that are 20 miles apart. They are both headed for the same intersection, as depicted in the diagram below. Car A is traveling at 30 mph, and Car B is traveling at 45 mph. Which car will reach the intersection first?

picture: https://imgur.com/a/Zh9d8ax

Part I: Determine how far Car B has to travel to reach the intersection.  

Part II: Determine the time necessary for each car to reach the intersection. Round your answers to the nearest hundredth of an hour.  

Part III: Which car reaches the intersection first, and by how many hours?  

Answer 1

A goes 30 mph, B 45 mph, both headed to the same intersection.

which gets there 1st??

construct a triangle angle A=95, b=18, c=20

use law of cosines to find missing 3rd side

a^2 = b^2 + c^2 -2bcCos95 = 18^2 + 400 - 720Cos95

a^2 = 724 -720Cos95 =about 724-525.72= 198.28

a = sqr198.28 = about 14.08

a = about 14.08 miles

now use distance formula d=rt = distance = rate of speed times time

t = d/r

for A, 14/30 = 7/15 = about .467 hours= about .47 hours = about 28 minutes

for B 18/45 = 2/5= .4.00 hours = 2/5 hrs = 24 minutes

2/5 < 7/15, so B gets there 1st as it takes less time than A to get to the intersection

B beats A by about 4 minutes = about 4/60 = 1/15 of an hour = about .07th of an hour

Answer 2

Velocity of car A = V_a = 30 mph

Velocity of car B = V_b = 45 mph

Need to use law of cosines.

a²=b²+c²-2abcos(A)

From the picture, A= 95°, b=18 and c=20

Part I

'a' is the distance car B needs to travel

a= √(b²+c²-2abcos(A))

Part II

Time for car A = T_a = 18/V_a

Time for car B = T_b = a/V_b

Part III

The car with the lowest time will get to the intersection first.

Find the difference between T_a and T_b

_