change in relative distance from d1 to d2 where d1>d2 Two cyclists start from the same place to ride in the same direction.A starts at noon with a speed of 8km/hr and B starts at 2pm with a speed of 10km/hr.At what times A and B will be 5km apart ?
My thought process:
As A starts early at 12 so it will have already covered 16km(8*2).
so S relative=V relative*t or say 16=2*t1 and thus t1=8.
Now we want Srelative to be 5 so 
5=2*t2 and t2=2 and a half hour so they will meet at t1-t2 .
Is this correct process ?
 A: $$x_1(t)=v_1(t-t_1)+x_0=8(t-0)$$
$$x_2(t)=v_2(t-t_2)+x_0=10(t-2)$$
thus the condition
$$|x_2(t)-x_1(t)|=|2t-20|=5$$
 gives two solutions
$$t=12,5  \text{  or } t=7,5$$
for example,
at $t=7,5$
$$x_1=8(7,5-0)=60 \; km$$
and
$$x_2=10(7,5-2)=55 \; km$$
A: I'll give an alternative, which might look a bit less tedious.
Assume $t=0$ is at 12 noon.
Since $A$ starts at $t=0$ and goes at a speed of 8, we can express the distance traveled by him as 
$$a(t) = 8t$$
But $B$ starts with a delay of 2 units; this means his graph shifts to the right by 2 units : 
$$b(t) = 10(t-2)$$
Now that equations are setup, you simply have  to solve $|b(t)-a(t)|=5$
A: By the time $B$ starts, $A$ is ahead by $16$ km. The velocity of $B$ with respect to $A, v_{BA}=v_B-v_A=2$ kmph. Therefore, the distance between $A$ and $B$ changes at $2$ kmph, and after $t$ hours from $B$'s departure, the distance between them is given by $|16-2t|$.
We want $|16-2t|=5\implies t=5.5, 10.5$ hours. Since $B$ started at $2$ pm, they are $5$ km apart at $5.5$ and $10.5$ hours after $2$ pm, that is, at $7:30$ pm, $12:30$ am.
