SOLUTION: A car goes 40 yards / sec. A bike goes 20 yards / sec. If they start at the same point, when do they meet? (I don't think they do, since the bike will never catch up to the car.

Question 1125555: A car goes 40 yards / sec.
A bike goes 20 yards / sec.
If they start at the same point, when do they meet?
(I don't think they do, since the bike will never catch up to the car.)
If the bike is given a 30 yard head start, at what time (seconds) will they meet? And at what distance?
I also know speed = distance / time.
Thank you!!!
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!
Part (i): You are correct, if they are moving in the same direction and the bike and car start at the same point.

Part (ii):
speed = distance / time �> distance = speed*time
d(bike) = 30yds + 20yds/sec*(t)
d(car) = 40yds/sec*(t)
They meet when d(car) = d(bike) ==> 30+20t = 40t, solving for t: and the distance from the car's starting point is 1.5s*40yds/s =
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I chose the car's starting point as the reference point, the "beginning." Thus, the 1.5*40 in the 2nd part ("where they meet") is the distance from where the car started, so I used the car's speed.

It would be just a valid (although a bit obscure?) to choose the bike's starting point, but remember the bike had a 30yd head start over the car. *If* one chooses the bike's starting point, then they meet at 1.5*20 = 30yds from the BIKE's starting point (and yes, you use the bike's speed in this case), and that 30yds for the bike corresponds to a distance traveled of 30+30 = 60yds for the car.

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