Velocity = distance/time
P=a = x/t1
Q=a+b = x/t2
R=a+2b = x/t3
Q stats p hours after P:
(a+b)=X/(t1+p)
hours after Q should R start:
a+2b = X/(t2+h)
t1=x/a
t2=t1+p
t3=t1+p+h
Velocity Q = R
by manipulating the velocity formula:
Velocity/time = distance
Q distance = R distance
Velocity*time=distance:
(a+b)/(t1+p) = (a+2b)/(t1+p+h)
solve for h:
P, Q and R start from the same place X at (a) kmph, (a+b) kmph and (a+2b) kmph respectively.
If Q starts p hours after P, how many hours after Q should R start, so that both Q and R overtake P at the same time?
2 answers
t1+p = t2
so:
(a+b)/t2 = (a+2b)/(t2+h)
h= ((a+2b)*t2)/(a+b) - t2
h=b*t2/(a+b)
so:
(a+b)/t2 = (a+2b)/(t2+h)
h= ((a+2b)*t2)/(a+b) - t2
h=b*t2/(a+b)