John travels 50 km/h, Mike travels 40 km/h. Travelling in a straight line, east to west, no stoplights and they leave at 10:00 a.m. and live 550 km apart.
1. How far from John's house do they meet?
2. What time do they meet?
2006-09-18 05:26:33 · 8 answers · asked by RefereeIV 1 in Science & Mathematics ➔ Mathematics
Since they are moving towards each other their relative velocity =
50 km/h + 40 km /h = 90 km /h
distance = 550 km
time = 550 / 90 = 6.11 hours
time of their meeting = 10:00 + 6 +( 0.11 x 60 ) = 4:06 pm
since john travelled for 6.11 hours
distance travelled by him = 50 x 6.11 = 305.5 km
2006-09-18 05:54:34 · answer #1 · answered by jazideol 3 · 1⤊ 0⤋
Set up a table to see where each of them after each hour:
Hour John's Distance Mike's Distance Sum
1 50 km 40 km 90
2 100 km 80 km 180
3 150 km 120 km 270
Continue until you get the sum to be 550...
2006-09-18 06:00:01 · answer #2 · answered by mathsmile 2 · 0⤊ 0⤋
Time travel 550 / (40 + 50) = 6.66666 = 6 hours and nearly 7 minutes.
Meeting time 16.07
John travel 50 * 6.66666 = 333.3333 km
2006-09-18 05:34:54 · answer #3 · answered by Anonymous · 1⤊ 0⤋
distance = speed * time
John: distance = 50t
Mike: distance = 550+40t
550+40t = 50t Solving this yields t = 55 hr THey meet 55 hours after 10 am
and they meet at a distance of (55*50=2750) km from John's House
2006-09-18 05:56:05 · answer #4 · answered by questionasker 2 · 0⤊ 1⤋
From the be conscious subject we get here equation: (Sa + Sb)T = D the place: Sa = velocity of practice a = 80 (MPH) Sb = velocity of practice b = 70 (MPH) T = Time (hr) D = Distance coated = 210 (Miles) understanding those variables we are able to replace in: (80 + 70)T = 210 Simplifying: (a hundred and fifty)T = 210 Divide the two components with the aid of "a hundred and fifty" (a hundred and fifty)T / a hundred and fifty = 210 / a hundred and fifty T = 210 / a hundred and fifty T = 1_4/10 Hr Or: T = one million hr. 24 min.
2016-10-15 03:20:03 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Man... the constant accelleration equations you learn in college physics would work wonders for this question.
2006-09-18 05:34:41 · answer #6 · answered by Jaques S 3 · 0⤊ 1⤋
FWIW, the answer by jazideol is correct and clear.
2006-09-18 07:23:34 · answer #7 · answered by Anonymous · 1⤊ 0⤋
do your own homework
2006-09-18 05:30:35 · answer #8 · answered by mytrollinid 5 · 0⤊ 1⤋