change in relative distance from d1 to d2 where d1>d2

Question

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 ?

Answer 1

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 $|b(t)-a(t)|=5$

Answer 2

$$x_1(t)=v_1(t-t_1)+x_0=8(t-0)$$ $$x_2(t)=v_2(t-t_2)+x_0=10(t-2)$$

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 $$x_2=10(7,5-2)=55 \; km$$

Answer 3

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.