Question 1125555
Part (i):  You are correct, if they are moving in the same direction and the bike and car start at the same point.<br>

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:  {{{highlight( t=1.5s ) }}} and the distance from the car's starting point is  1.5s*40yds/s = {{{ highlight( 60yds )}}}

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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. <br>

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.