Question

2 cars start at Heights high and ride in opposite directions. The first driver is traveling 15 mphfaster than the second driver. After 4 hours the cars are 415 miles apart. Find the speed of eachdriverFormat your answer like this ---> 1st driver. x mph, 2nd driver: y mph

Answer 1

`\begin{gathered} 1stdriver.\rightarrow58.125\text{ mph} \\ 2nddriver.\rightarrow43.125\text{ mph} \end{gathered}`

Step-by-step explanation

Step 1

time= 4 hours

total distance = x+y=415

`x+y=415\text{ Equation (1)}`

so,for car 1

`\begin{gathered} \text{speed}=\frac{dis\tan ace\text{ }}{\text{time}} \\ \text{spped}\cdot\text{time}=\text{ distance} \\ v_1\cdot4\text{ hours=x} \\ 4v_1=x\rightarrow Equation(2) \end{gathered}`

Now, for car 2

`\begin{gathered} \text{speed}=\frac{dis\tan ace\text{ }}{\text{time}} \\ v_2=(y)/(4) \\ v_2=v_1-15,\text{then} \\ v_1-15=(y)/(4) \\ 4v_1-60=y\text{ Equation (3)} \end{gathered}`

Step 2

solve the equations

`\begin{gathered} x+y=415\text{ equation (1)} \\ 4v_1=x\rightarrow Equation(2) \\ 4v_1-60=y\text{ Equation (3)} \end{gathered}`

a) replace x value from equation (3) and (2) in equation (1)

`\begin{gathered} \\ x+y=415\text{ equation (1)} \\ 4v_1+4v_1-60=415 \\ 8v_1=415+60 \\ 8v_1=465 \\ \text{divide both sides by 8} \\ (8v_1)/(8)=(465)/(8) \\ v_1=58.125\text{ mph} \end{gathered}`

so, the speed of the first car is 58.125 mpj, now replace the v1 , to find v2

The first driver is traveling 15 mph faster than the second driver

`\begin{gathered} v_1=v_2+15 \\ 58.125=v_2+15 \\ \text{subtract 15 in both sides} \\ 58.125-15=v_2+15-15 \\ 43.125=v_2 \end{gathered}`

Hence

`\begin{gathered} 1stdriver.\rightarrow58.125\text{ mph} \\ 2nddriver.\rightarrow43.125\text{ mph} \end{gathered}`

I hope this helps you