# 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 ?

• Your method looks perfect! But $8-2.5 = 5.5$ is the number of hours from 2pm that they will be 5 km away, right ? This means they will be 5 km away precicely at $7:30PM$ – rsadhvika Dec 10 '18 at 19:58
• The answer is $t_1\pm t_2$ hours ($A$ ahead/behind $B$ by $5$ km) after $2$ pm, since you measured $t$ hours from the departure of $B$. – Shubham Johri Dec 10 '18 at 21:40

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

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

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

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.