I need some math help!?

Question

How do you go about solving this problem. Thanks in advance!

Two cyclists start biking from a trail’s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

Answer 1

theheadstart of the first cyclist=6*3=18 miles
relativespeed of the second cyclist=10-6=4 mph
time taken toclosethe distance=s/v=18/4=4 1/2 hrs

Answer 2

Set up an equation to describe the position of each cyclist at time t, starting from time 0. The first cyclist is traveling at 6 mph and starts at time 0, so her position is just x1 = 6t, with time in hours and x in miles. The second cyclist travels at 10 mph, but starts at time t = 3, which means his position is x2 = 10(t - 3). Since you need to know when the second cylclist passes the first, you need to find the time t at which their positions are equal. So x1 = x2 ==> 6t = 10(t - 3), which is a straightforward algebra problem. This will give you the time that elapses from time 0. You need to subtract 3, because you've been asked to find the time that elapses after time t = 3 when the second cyclist starts.

Answer 3

I suggest this is a d=rt problem. in 3 hours the first cyclist is 18 miles down the road. T second goes at 10mph and is gaining at 4 mph. At 4 mph, how long to go the 18 miles between?
t=d/r=18/4=4.5 hrs. Is it all clear?

Answer 4

y=6x + 18 (got this by 3 hours * 6miles/hour)
y=10x

set two equations equal

6x+18=10x
18=4x
x=18/4 (this is the amount of time it will take after the 3 hours have passed for them to meet).

Answer 5

hey! i remeber doing this sum.but i am so sorry i frgt how to do it.which class r u in by the way?i am in the 7th.i hav harder sums than these.email me [e-mail]