At 2:00 PM, a car leaves town A and another leaves town B.The distance between the two towns is 60 km. One car travels 10 kph faster than the other. The car pass each other at 2:30 PM. How far does each travel?

Guest Jun 10, 2020

#1**+1 **

**At 2:00 PM, a car leaves town A and another leaves town B. The distance between the two towns is 60 km. One car travels 10 kph faster than the other. The car pass each other at 2:30 PM. How far does each travel?**

\(\begin{array}{|rcl|c|rcl|} \hline &(1)& && &(2) \\ \hline v_1 &=& \dfrac{s_1}{t} && v_2 &=& \dfrac{s_2}{t} \\ && & \boxed{v_1 = v_2+10,\\ s_1+s_2=60,\ s_2 = 60-s_1 }\\ v_2+10 &=& \dfrac{s_1}{t} && v_2 &=& \dfrac{60-s_1}{t} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)-(2): & v_2+10-v_2 &=& \dfrac{s_1}{t}-\dfrac{60-s_1}{t} \\\\ & 10 &=& \dfrac{s_1}{t}-\dfrac{60-s_1}{t} \quad | \quad *t \\\\ & 10t &=& s_1 - (60-s_1) \\ & 10t &=& 2s_1 - 60 \\ & 2s_1 &=& 10t+60 \quad | \quad :2 \\ & \mathbf{s_1} &=& \mathbf{5t+30} \\ \hline & s_2 &=& 60-s_1 \\ & s_2 &=& 60-(5t+30) \\ & \mathbf{s_2} &=& \mathbf{30-5t} \\ \hline \end{array}\)

\(\text{Let $t=2:30\ PM-2:00\ PM=0.5$ hours}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_1} &=& \mathbf{5t+30} \quad | \quad t=0.5 \\ s_1 &=& 5*0.5+30 \\ \mathbf{s_1} &=& \mathbf{32.5\ \text{km} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{s_2} &=& \mathbf{30-5t} \quad | \quad t=0.5 \\ s_2 &=& 30-5*0.5 \\ \mathbf{s_2} &=& \mathbf{27.5\ \text{km} } \\ \hline \end{array}\)

**How far does each travel:**

One car travel \(\mathbf{32.5\ \text{km}}\) and the other car travel \(\mathbf{27.5\ \text{km}}\)

heureka Jun 10, 2020

#2**+1 **

Car A = x_{a}

Car B = x_{a}+10

rate x time = distance time is 1/2 hr

(x_{a} + xa+10)* 1/2 = 60

x_{a} = 55 k/hr in 1/2 hr car A travels 27.5 km

x_{a}+10 = 65 k/hr in 1/2 hr car B travels 32.5 km Check: 27.5 + 32.5 = 60 km

ElectricPavlov Jun 10, 2020