MHB Calculating Speed: Solving a Distance and Time Problem

[Marcelo Arevalo]

Distance & Speed Problem

City B is between City A and City C.
Ken is riding a bike from B to C, a distance of 16 km.
After he has gone 6 km, Mark begins to drive a car at 60 km per hour from A to C.
If Ken continues onward, he will reach C at the same time as Mark. If he turns back, he will reach B at the same time as Mark. What is his speed, in km per hour?

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Given :

Distance;
A ------------------ B ------------------- C

Ken is at B going to C with distance of 16 km.
B ----X-----------C

B to X is 6 km. also X to C is 10 km.

Now I am stuck.. please help.

 

[MarkFL]

I don't think we can assume $B$ is midway between $A$ and $C$. Let's let:

$$v_K$$ = Ken's speed

$$d_{AB}$$ = distance from $A$ to $B$

$$t_{K_1}$$ = time traveled by Ken if he turns back.

$$t_{M_1}=t_{K_1}-\frac{6}{v_K}$$ = time traveled by Mark, if Ken turns back.

$$t_{K_2}=\frac{4}{3}t_{K_1}$$ = time traveled by Ken if her continues.

$$t_{M_2}=t_{K_2}-\frac{6}{v_K}=\frac{4}{3}t_{K_1}-\frac{6}{v_K}$$ = time traveled by Mark, if Ken continues.Now, if Ken turns back then he has traveled 12 km and Mark has traveled $d_{AB}$...so we may write:

$$12=v_Kt_{K_1}\tag{1}$$

$$d_{AB}=60\left(t_{K_1}-\frac{6}{v_K}\right)\tag{2}$$

If Ken continues, then we may write:

$$16=\frac{4}{3}v_{K_1}t_{K_1}$$ (we have no new information here)

$$d_{AB}+16=60\left(\frac{4}{3}t_{K_1}-\frac{6}{v_K}\right)\tag{3}$$

We now have 3 equations and 3 unknowns...can you proceed?

 

[Greg]

Alternatively,

Let $y$ be the distance from city A to city B. Let $t_1$ be the time it takes Mark to travel from city A to city B. Let $M$ be Mark's speed and let $K$ be Ken's speed. So we have

$$M\cdot t_1=y$$
$$K\cdot t_1=6$$

Dividing these two equations gives $\dfrac MK=\dfrac y6$.

Let $t_2$ be the time it takes Mark to travel from city A to city C. Then we have

$$M\cdot t_2=y+16$$
$$K\cdot t_2=10$$

Dividing these two equations gives $\dfrac MK=\dfrac{y+16}{10}$, so we have

$$\dfrac y6=\dfrac{y+16}{10}\implies y=24\text{ km.}$$

You should now be able to complete the problem.

 

[Marcelo Arevalo]

From your equation 1 of MarkFL post :

turn-around travel would be 12 km.

Ken on Bike; B turn Aroound to B
12/15 -->> 15km/hr assume velocity
12/15 = 0.8 hr

if he continues to the end;
16/15 = 1 1/15 hr

Mark on Car:
using Ken's end time = 1 1/15 hr X 60 kph = 64 km

A to B = 64 Km - 16 Km (B to C) = 48 Km
So: A to B
48 / 60 = 0.8 hr (Same time as Ken's Turn Around)

A to C = 48 + 16 = 64 Km

Then: 64 / 60 = 1 1/15 hr

therefore Ken's Speed is 15 Kph

 

[Marcelo Arevalo]

Sorry I am used to a trial & error solution. I am practicing to solve mathematics problem using the algebra solutions. that's why I keep on asking for your help. so I cam polished my thoughts and put it into a detailed solutions.

 

[MarkFL]

Yes, I also get 15 kph for Ken...here are the 3 equations I gave:

$$12=v_Kt_{K_1}\tag{1}$$

$$d_{AB}=60\left(t_{K_1}-\frac{6}{v_K}\right)\tag{2}$$

$$d_{AB}+16=60\left(\frac{4}{3}t_{K_1}-\frac{6}{v_K}\right)\tag{3}$$

If we solve (2) and (3) for $d_{AB}$, and equate the results, we find:

$$60\left(t_{K_1}-\frac{6}{v_K}\right)=60\left(\frac{4}{3}t_{K_1}-\frac{6}{v_K}\right)-16$$

Divide through by 60:

$$t_{K_1}-\frac{6}{v_K}=\frac{4}{3}t_{K_1}-\frac{6}{v_K}-\frac{4}{15}$$

Add $$-t_{K_1}+\frac{6}{v_K}+\frac{4}{15}$$ to both sides:

$$\frac{4}{15}=\frac{1}{3}t_{K_1}$$

Multiply through by 3:

$$\frac{4}{5}=t_{K_1}$$

And so (1) tells us:

$$v_K=\frac{12}{\dfrac{4}{5}}=15$$

However, like greg1313, I get a distance of 24 km from A to B.

 

[Marcelo Arevalo]

Using Greg1313 :

A------------B----X-----C

AB = 24 km + BC = 16 km
t = 40 / 60 = 2/3 hr

XC = 10 km
v = 10 km / 2/3 hr = 15 Kph

To Check :

AC = 40 / 60 = 2/3 hr
XC = 10 / 15 = 2/3 hr

AB = 24/60 = 2/5 hr
XB = 6/15 = 2/5 hryes.. the speed of Ken's bike is 15 Kph.Thank you all !