Arthur D. answered 12/03/14
Tutor
4.9
(274)
Mathematics Tutor With a Master's Degree In Mathematics
distance=rate*time
walker A walks 1.5 m more than walker B because A starts at the 0.5 m mark and B starts at the 2 m mark (2-0.5=1.5)
d+1.5=1*t for walker A
d=0.5*t for walker B
0.5t+1.5=t
1.5=t-0.5t
1.5=0.5t
15=5t
t=15/5=3 seconds elapses when walker A catches up to walker B
d=0.5*3
d=1.5 m
walker B walks 1.5 m starting at the 2 m mark and ends up at the 3.5 m mark when walker A catches him
walker A starts at the 0.5 m mark and walks for 3 seconds at a speed of 1 m/sec, therefore, walker A walks 3 m starting at the 0.5 m mark and ends up at the 3.5 m mark where again he catches up to walker B