Two riders are in a bike race. The second rider starts half an hour late but rides 4 miles per hour faster than the first rider. If they both finish 3 hours after the race starts, how fast are they each riding?

Question

anonymous · Answer

Let's call x the amount of time the first rider has been on the road.  Let's also call v the velocity of the first rider.  An equation for the distance for the first rider has traveled is
$$D_1=vx$$
The second rider has an initial start time of (x-0.5) because he starts half an hour later than the first rider.  The second rider's velocity is 4 mph faster than the first rider, or (v+4).  This gives us an equation for the distance the second rider has traveled
$$D_2=(v+4)(x-0.5)$$
They finish 3 hours after the race starts (x=3), which also means they traveled the same distance, so
$$D_1=D_2$$$$vx=(v+4)(x-0.5)$$$$v(3)=(v+4)(3-0.5)$$$$3v=(v+4)2.5$$Distribute
$$3v=2.5v+10$$Subtract 2.5v from both sides to get v on one side.
$$0.5v=10$$Multiply by 2 to get 1v
$$v=20$$
The second rider is riding 4 mph faster, so the second rider is riding 24 mph.

rosey · Answer

i can't help but u should b a fan of lil ol me :)

anonymous · Answer

Who should?