Uniform Motion Related Problem

Question

Chester drove his car at 8:00 a.m. at speed of 45 km/hr. At 9:15a.m. Michael followed with his car. What should be Michael's speed so as to be able to overtake Chester at 1:00 p.m.? How far has Chester travelled when Michael overtook him?

Answer 1

9:15 - 8:00 = 1h and 15 min. = 1.25h head-start.

d = r1*t = 45km/h * 1.25h = 56.25 km
Head-start.

r2*t = r1*t + 56.25

t = 9:15AM to 1PM = 3h and 45min = 3.75h

r2 * 3.75 = 45*3.75 + 56.25
r2*3.75 = 225
r2 = 60km/h = Michael's speed.

r1*t + 56.25 = 45*3.75 + 56.25 = 225 km = Chester's distance when overtaken by
Michael.