Frank started travelling 45 mph. When Frank was 1/3 miles away, Daniel started from the same point at 50 mph to catch up with Frank. How long will it take Daniel to catch up to Frank?

Step-by-step explanation:

Initial velocity of frank=45mph,

Now frank at a distance 1/3miles

Daniel started with a speed of 50mph,

Let Daniel catch up with Frank at time t

Speed = distance/time

Time=distance/speed

For frank, he has already traveled for a time of

t, =d/s

t=1/3÷50

t=1/150hour

Let now assume they meet at distance x,

Time for Daniel to reach x

t=d/s

t=x/50. Equation 1

Time for Frank to reach x, he has already traveled 1/3 miles, then he has x-1/3miles to travelled

t=d/s

t=(x-1/3)÷45. Equation 2

Then both time are same, let equate them

time frank =time Daniel

x/50=(x-1/3)/45

45x=50(x-1/3)

45x=50x-50/3

45x-50x=-50 /3

-5x=-50/3

x=50/3÷5

x=10/3

x=3.333miles

They will meet at 3.3333miles, not to know the time, let substitute into equation 1

t=x/50

t=3.33333/50

t=0.0667hours

Or t=4secs

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

answer: im gonna go with milli

step-by-step explanation: